Exercise : Solution of Questions on page Number : 156
Answer: 1
Yes, the centre of mass can be at a point outside the body.
For example, the centre of mass of a ring lies at its centre, which is not a part of the ring.
Answer: 2
Yes, if all the particles of a system lie in the X–Y plane, then it’s necessary that its centre of mass lies in the X–Y plane.
zcm=m1z1+m2z2+...mnzn∑m
As all the particles lie in the X–Y plane, their z-coordinates are zero.
Therefore, for the whole system, z cm= 0; i.e., its centre of mass lies in the X–Y plane.
Answer: 3
Yes. As a cube is a 3-dimensional body, all the particles of a system lying in a cube lie in the x,y and zplane.
Let the ith element of mass ∆mi is located at the point (xi,yi,zi).
The co-ordinates of the centre of mass are given as:
X=1M∑i=ni=1(∆mi)xiY=1M∑i=ni=1(∆mi)yiZ=1M∑i=ni=1(∆mi)zi
X, Y and Z lie inside the cube because it is a weighted mean.
Answer: 4
(a) Yes
Consider a charge distributed in X-Y plane.•••••••••••••••
Xcm=−6q×0+q×5a−6q+q=−a
(b) Yes. Because the z-coordinates of all the charges are zero, the centre of charge lies in X-Y plane.
(c) No, it is not necessary that the centre of charge lies in the cube because charge can be either negative or positive.
Answer: 5
In order to simplify the situation, we consider that the weight Mg of an extended body acts through its centre of mass.
Although the earth attracts all the particles, the net effect can be assumed to be at the centre of mass.
Answer: 6
There is no violation of conservation of momentum because in the earth’s frame the component of tension is acting in the horizontal direction.
Answer: 7
The trunk does not recoil as the girl jumps off on the platform because the force exerted by the girl is less than the limiting friction between the platform and the iron trunk.
Answer: 8
Yes.
For example, consider particle-1 at a velocity of 4 ms-1 and particle-2 at a velocity of 2 ms-1 undergo a head-on collision.
The velocity of particle-1 decreases but particle-2 increases. Therefore, at an instant, their velocities will be equal.
Answer: 9
No. As the collision experiment is being done on a horizontal table in the elevator that is accelerating up or down in vertical direction, no extra force is experienced in the horizontal direction. Hence, the objects in the horizontal direction remain unaffected.
Answer: 10
No, due to the non-inertial character of the frame and the presence of a pseudo force, both the equations, i.e., Velocity of separation = Velocity of approach and Final momentum = Initial momentum, do not remain valid in the accelerating car.
Answer: 11
No. As the potential energy can have a negative value, the total energy of the system may sum up to zero.
For example:
Two masses A and B having masses 2 kg and 4 kg respectively move with a velocity of 4 ms-1 in opposite directions.
Kinetic energy of system (A and B)
=12×2×42 +12×4×42 =48 J
If the gravitational potential energy of the system is −48 J, the total energy of the system will be zero.
However, the linear momentum will be non-zero.
Answer: 12
Yes, the kinetic energy of the particle can be determined if the value of linear momentum is known.
The kinetic energy is calculated using the formula:
K.E=12mv2=p22mwhere, p is the linear momemtum having value mv.
But linear momentum cannot be determined even if the kinetic energy is known because linear momentum is a vector quantity, whereas kinetic energy is a scalar quantity. Thus, the direction of the linear momentum remains unknown, however its magnitude can be calculated.
Answer: 13
The distance of centre of mass of a uniform hemisphere from its centre will be less than r/2 because the portion of the hemisphere lying below r/2 from the diameter is heavier than the portion lying above r/2.
Answer: 14
More effort is needed when the cage is closed, while less effort is required when the cage has rods to let the air pass. When a bird flies from its position, it pushes the air downwards. Thus, when the bird is in a cage, the net downward force will be equal to the weight of the cage plus the downward force due to air (the weight of the bird).
However, if the cage has rods to let air pass, the downward force exerted by air become less. Therefore, less effort will be required to hold the cage.
Exercise : Solution of Questions on page Number : 157
Answer: 15
According to the question, the weight of plank is very less as compared to the fat person. Therefore, the centre of mass of the whole system effectively lies on the person. As the net external force on the system is zero, the centre of mass of the system does not move.
Answer: 16
From the figure, it can be seen that when a high-jumper successfully clears the bar, it is possible that her centre of mass crosses the bar from below it because the legs as well as the arms of the high-jumper are below the bar.
Hence, the point shown in the figure can be her centre of mass.
Answer: 17
The person shown on the right hand side of the figure is more likely to fall down because in the given cart frame the pseudo force will be in backward direction.
Answer: 18
It is not necessary that the linear momentum of a system remains constant even if no external force acts on it because during collision, the sum of magnitudes of momenta does not remain constant.
Answer: 19
Yes, if the external force applied on the particle is zero, its speed does not change and hence, the momentum remains constant.
Answer: 20
Yes, it’s proper to say that the force generated by the engine accelerates the car. When petrol burns inside the engine, the piston moves, which in turn rotates the wheel. As the wheel rotates, the frictional forces from the road moves the car.
Answer: 21
The frictional force acting between the surface of the table and the ball is responsible for the change in momentum of the ball. As the force opposes the motion of the ball, it stops after moving some distance.
Answer: 22
Considering the table plus the ball as a system, it can be said that the frictional force is responsible for the change in the momentum. As the force acts between the surface of the/images/hc-verma/Physics//Class12-physics-chapter-9-157-6.1.png/images/hc-verma/Physics//Class12-physics-chapter-9-157-6.1.png table and ground, it opposes the motion of the table plus the ball. Hence, the ball slows down and the momentum of the system decreases.
Answer: 23
In view of the principle of conservation of momentum, the given situation is possible because as a beta particle is ejected, another particle called an antineutrino is also ejected.
Answer: 24
According to the question, the van is standing on a frictionless surface. When throwing something in backward direction, the persons sitting inside the van sets the van in motion in the forward direction according to the principle of conservation of linear momentum.
Answer: 25
No. If the masses are different, the velocities in a one-dimensional collision cannot be interchanged because that would be violation of the principle of conservation of momentum.
Answer: 1
(c) A is correct but B is wrong
In a non-inertial frame, the position of centre of mass of the particle does not change but an additional pseudo force acts on it.
Answer: 2
/images/hc-verma/Physics//Class12-physics-chapter-9-157-6.1.png/images/hc-verma/Physics//Class12-physics-chapter-9-157-6.1.png(d) B implies A but A does not imply B.
The centre of mass of a system is given by,
R→=1M∑mir→i
On differentiating the above equation with respect to time, we get:
dR→dt=1M∑midr→idt
As the centre of mass of the system remains at rest, we have:
1M∑midr→idt=0∑miv→i=0
This implies that the linear momentum of the system remains constant.
Answer: 3
(d) B implies A but A does not imply B.
If the linear momentum of a system is zero,
⇒m1v→1 + m2v→2 + … = 0
Thus, for a system of comprising two particles of same masses,
v→1=-v→2 …(1)
The kinetic energy of the system is given by,
K.E.=12mv→12+12mv→22
Using equation (1) to solve above equation, we can say:
K.E.≠0
i.e. A does not imply B.
Now,
If the kinetic energy of the system is zero,
⇒12mv→12 + 12mv→22 = 0
v1=±v2
On calculating the linear momentum of the system, we get:
P→= mv→1 + mv→2taking v1=-v2, we can write:P→= 0
Hence, we can say, B implies A but A does not imply B.
Answer: 4
(d) both A and B are false.
As the velocity of the particle depends on the frame of reference, the linear momentum as well as the kinetic energy is dependent on the frame of reference.
Answer: 5
(b) ≤ R
Distance of the centre of mass from the origin is given by,
R’ = 1M∑i=1nmir→i
Let half of the particles lie on +Y-axis and the rest of the particles lie on +X-axis.
YCM = m3Rj→+ m4Rj→m1 + m2 + m3 + m4 +…= Rj→2
Similarly, XCM = Ri→2
Therefore, the coordinates of centre of mass are R2,R2.
Distance of the centre of mass from the origin=R2
For the given situation, R'<R.
In general, R’ ≤ R.
Answer: 6
(b) inside the square plate
Let m1 be the mass of circular plate and m2 be the mass of square plate.
The thickness of both the plates is t.
mass = density × volumem1 = ρπd22t m2 = ρd2t
Centre of mass of the circular plate lies at its centre.
Let the centre of circular plate be the origin.
r→1=0
Centre of mass of the square plate lies at its centre.
r→2=2dNow,R = m1r→1 + m2r→2m1 + m2 = m1 × 0 + ρd2t × 2dρπd22t + ρd2t = 2dπ4 + 1 = 1.12d⇒R>d
∴ Centre of mass of the system lies in the square plate.
Exercise : Solution of Questions on page Number : 158
Answer: 7
(b) 12a→
Acceleration of centre of mass of a two-particle system is given as,
a→cm=m1a→1+m2a2→m1+m2 …(1)
According to the question,
m1=m2=ma1=0a2=a
Substituting these values in equation (1), we get:
a→cm = m × 0 + ma→2m = 12a→
Answer: 8
(b) the kinetic energy but not the linear momentum
Internal forces can not change the position of centre of mass of a system. Therefore, linear momentum of the system is constant, whereas kinetic energy of the system is not.
Answer: 9
(c) gravitational potential energy of the block
When the block kept at rest is hit by a bullet, the block acquires certain velocity by the conservation of linear momentum.
Therefore, the linear momentum and the kinetic energy of the block change.
As some of the kinetic energy carried by the bullet transforms into heat energy, its temperature also changes.
However, the gravitational potential energy of the block does not change, as the height of the block does not change in this process.
Answer: 10
(d) is independent of h.
As the uniform sphere is placed on a smooth surface, the sphere only slips.
The acceleration of the centre of the sphere is given by,
Applied force Mass of the sphere ,which is independent of height h.
Answer: 11
(c) does not shift horizontally
As the body falls vertically downwards, no external force acts in the horizontal direction.
Hence, the centre of mass does not shift horizontally.
Answer: 12
(b) of the box plus the ball system remains constant
Consider the box and the ball a system. As no external force acts on this system, the velocity of the centre of mass of the system remains constant.
Answer: 13
(c) in opposite directions with equal speeds
Using the principle of conservation of linear momentum, we can write:
Initial momenta = Final momenta
⇒ M × 0 = M2v1 + M2v2⇒v1 = -v2
where M is the initial mass of the body, at rest.
The final mass of the two pieces moving with equal speeds in opposite direction is equal to M2.
Answer: 14
(c) 2mv2r must be acting on the system
Total mass of the system = 2m
To move the centre of the system in a circle of radius r with a uniform speed v, the external force required is 2mv2R.
Answer: 15
(d) momentum, but neither kinetic energy nor temperature
Linear momentum of a system remains constant in a collision. However, the kinetic energy and temperature of the system may vary, as their values depend on the type of collision.
Answer: 16
(d) m1v1→+m2v2→ must be parallel to v→
By the law of conservation of linear momentum, we can write:
Initial momentum = Final momentum⇒mv→ = m1v→1 + m2v→2⇒(m1v→1 + m2v→2) must be parallel to v→
Answer: 17
(a) 3V cos θ
The linear momentum is conserved in horizontal direction.
∴ Initial momentum = Final momentum
⇒mv cos θ = -m2v cos θ + m2v’⇒v’ = mv cos θ
Answer: 18
(a) the initial kinetic energy is equal to the final kinetic energy
As no energy is lost into heat in an elastic collision, the initial kinetic energy is equal to the final kinetic energy.
Answer: 19
(b) the final kinetic energy is less than the initial kinetic energy
As some energy is loss into heat in an inelastic collision, the final kinetic energy is less than the initial kinetic energy.
Answer: 1
None.
The centre of mass of a system of particles depends on the product of individual masses and their distances from the origin.
Therefore, we may say about the given statements:
(a) Distance of particles from origin is not known.
(b) Masses are same but the distance of particles from the origin is not given.
(c) Distance of particles from origin is not given.
(d) It is not necessary that least one particle lies on the negative X-axis. The particles can be above the negative X-axis on X-Y plane.
Exercise : Solution of Questions on page Number : 159
Answer: 1
Taking BC as the X-axis and point B as the origin, the positions of masses m1 = 1 kg, m2 = 2 kg and m3 = 3 kg are
0,0, 1,0 and 12,32 respectively.
The position of centre of mass is given by:
Xcm, Ycm = m1x1 + m2x2 + m3x3m1 + m2 + m3, m1y1 + m2y2 + m3y3m1 + m2 + m3
= 1 × 0 + 2 × 1 + 3 × 121 + 2 + 3, 1 × 0 + 2 × 0 + 3 × 321 + 2 + 3= 712 m, 34 m
Answer: 2
Let OX be the x-axis, OY be the Y-axis and O be the origin.
Mass of O atom, m1= 16 unit
Let the position of oxygen atom be origin.
⇒x1=y1=0Mass of H atom, m2 = 1 unitx2 = -0.96 × 10-10 sin 52°my2 = -0.96 × 10-10 cos 52°Now, m3 = 1 unitx3 = 0.96 × 10-10 sin 52°my3 = -0.96 × 10-10 cos 52°The X coordinate of the centre of mass is given by : xcm = m1x1 + m2x2 + m3x3m1 + m2 + m3=16×0 + 1×-0.96×10-10 sin 52° + 1×0.96×10-10 sin 52° 1 + 1 + 16 = 0 The Y coordinate of the centre of mass is given by : ycm = m1y1 + m2y2 + m3y3m1 + m2 + m3 = 16×0 + 2×0.96×10-10 cos 52°1 + 1 + 16= 2×0.96×10-10 cos 52°18 = 6.4×10-12 m
Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is (0, 6.4×10-12 m).
Answer: 3
Let OX be the X-axis and point O (0, 0) be the origin of the system.
The mass of each brick is M.
The length of each brick is L.
Each brick is displaced with respect to another in contact by a distance L10.
∴ The X-coordinate of the centre of mass is given as,
Xcm= 17 mL10mL2+mL2+L10+………+mL2Xcm= L2+L2+L10+L5+L2+3L10+L2+L5+L2+L10+L27= 7L2+5L10+2L57 = 22L35
The X-coordinate of the centre of mass relative to the origin is 22L35.
Answer: 4
Let the centre O (0, 0) of the bigger disc be the origin.
Radius of bigger disc = 2R
Radius of smaller disc = R
Now,
m1 = μR × T × ρ
m2 = μ (2R)2 × T × ρ
where T is the thickness of the two discs, and
ρ is the density of the two discs.
Position of the centre of mass is calculated as:
x1=R, y1=0 x2=0, y2 =0
= m1x1+ m2x2m1+ m2, m1y1+ m2y2m1+ m2= πR2TρR + 0μR2TρR + μ(2R)2Tρ, 0m1+ m2= πR2TρR5μR2Tρ ,0 = R5, 0
⇒ The centre of mass lies at distance R5 from the centre of bigger disc, towards the centre of smaller disc.
Hence, the centre of mass of the system is R5,0.
Answer: 2
(c) may be all non-negative
(d) may be positive for some cases and negative for other cases
According to the question, the centre of mass is at origin.
∴ X = m1x1 + m2x2 + m3x3 +…m1 + m2 + m3 +……. = 0
⇒ m1x1 + m2x2 + m3x3 +… = 0
From the above equation, it can be concluded that all the x-coordinates may be non-negative.
In other words, they may be positive for some cases and negative for others.
Answer: 3
(a) the density continuously increases from left to right
(b) the density continuously decreases from left to right
As the density continuously increases/decreases from left to right, there will be difference in the masses of rod that lie on either sides of the centre of mass. Thus, the centre of mass of a rod in such a case will certainly not be at its centre.
Answer: 4
(b) must not accelerate
(c) may move
If the external force acting on a system has zero resultant,
then, acceleration of centre of mass = F→netM=0.
However, it may move uniformly with constant velocity.
Answer: 5
(b) v0 = 0, a0 ≠ 0
(d) v0 ≠ 0, a0 ≠ 0
If a non-zero external force acts on a system of particles, it causes the centre of mass of the system to accelerate with acceleration a0 at any instant t. In such a case, the velocity of centre of mass of the system of particles is either v0 or zero.
Answer: 6
(d) is equal to g
The acceleration of the centre of mass of two balls having masses m1 and m2 is given by
acm = m1a→1 + m2a→2m1 + m2 = m1g + m2gm1 + m2= g
Answer: 7
(a) the total momentum must be conserved
(d) the total kinetic energy must change
As no external force acts on the block, the linear momentum is conserved.
Some energy is used to break the block, thus the total kinetic energy must change.
Answer: 8
(b) the linear momentum remains constant
(c) the final kinetic energy is equal to the initial kinetic energy
(d) the final linear momentum is equal to the initial linear momentum.
During an elastic collision, all of the above statements are valid.
Answer: 9
(c) the total momentum of the ball and the earth is conserved
(d) the total energy of the ball and the earth remains the same
As the ball rebounds after hitting the floor, its velocity changes.
i.e. Velocity of ball before collision ≠ Velocity of ball after collision
Therefore, the momentum of the ball just after the collision is not same as that just before the collision.
The mechanical energy of the ball also changes during the collision.
However, the total momentum of the system (earth plus ball) and the total energy of the system remain conserved.
Answer: 10
(b) both the bodies move after collision
(c) the moving body comes to rest and the stationary body starts moving
By using the law of conservation of linear momentum we can write:
Initial momentum of the two bodies = Final momentum of the two bodies
⇒pi =pf⇒m1u = m1v1 + m2v2⇒v1 = v2 ≠ 0However, it may be possible that v1 becomes 0 and v2 becomes m1um2.
Answer: 11
All of the above.
(a) the velocities are interchanged
(b) the speeds are interchanged
(c) the momenta are interchanged
(d) the faster body slows down and the slower body speeds up
Exercise : Solution of Questions on page Number : 160
Answer: 5
Let O be the origin of the system (smaller disc plus bigger disc).
The density of the rods is ρ.
The thickness of rods is T.
Let m1 be the mass and R be the radius of the smaller disc, and
m2 be the mass and 2R be the radius of the bigger disc.
According to the question, the smaller disc is cut out from the bigger disc.
From the figure given below, we can write:
m1 = πR2Tρ Position of centre of mass of smaller disc,x1 = R, y1 = 0m2 = π(2R)2Tρ Position of centre of mass of bigger disc,x2 = 0, y2 = 0
Therefore, the position of centre of mass of the system,Xcm = m1x1 + m2x2m1 + m2, Ycm = m1y1 + m2y2m1 + m2
Since the smaller disc is removed from the bigger disc, the mass of the smaller disc is taken as negative (- 1).Xcm=-πR2TρR + 0m2-πR2Tρ + π(2R)2Tρ, Ycm=0 + 0m1 + m2Xcm=-πR2TρR3πR2Tρ, Ycm=0Xcm=-R3, Ycm=0
Hence, the centre of mass of the system lies at distance R3 from the centre of bigger disc, away from centre of the hole.
Answer: 6
Let m be the mass per unit area of the square plate and the circular disc.
⇒Mass of the square plate, M1 = d2m
Mass of the circular disc, M2 = πd24m
Let the centre of the circular disc be the origin of the system.
⇒ x1 = d, y1 = 0
x2 = 0, y2 = 0
⇒ Position of the centre of mass of circular disc and square plate:
=m1x1 + m2 x 2m1+m2,
m1y1 + m2y2m1 + m2= d2md + πd2/4 m×0d2m + π d2/4 m,
0 + 0m1 + m2= d3md2m(1 + π/4),
0= 4dπ + 4, 0
Hence, the new centre of mass of the system (circular disc plus square plate) lies at distance 4d(π+4) from the centre of circular disc, towards right.
Answer: 7
From the figure, the velocities of different masses can be written as:
For m1= 1.0 kg,
Velocity, v→1= -1.5 cos 37°i^-1.5 sin 37°j^= -1.2i^-0.9j^For m2= 1.2 kg,
Velocity, v→2= 0.4 j→For m3=1.5 kg,
Velocity, v→3= -1.0 cos 37°i^+1.0 sin 37°j^For m4=0.50 kg,
Velocity, v→4= 3.0 i^For m5=1.0 kg,
Velocity,v→5=2.0 cos 37° i^-2.0 sin 37°j^(cos 37°=45 and sin 37°=35)Vcm= m1v→1 + m2v→2 + m3v→3 + m4v→4 + m5v→5m1 + m2 + m3 + m4 + m5=11.0 + 1.2 + 1.5 + 1.0 + 0.501.0-1.5 × 45i→ – 1.5 × 35j→+…-2.0 × 35j→
On solving the above equation, we get:
Vcm is 0.20 m/s , at 45° below the direction, towards right.
Answer: 8
Let the two masses m1 and m2 be placed on the X-axis.
It is given that:
m1 = 10 kg
m2 = 20 kg
The first mass is displaced by a distance of 2 cm.
∴ Xcm= m1x1 + m2x2m1 + m2⇒Xcm= 10 × 2 + 20×230
As the position of the centre of mass remains unchanged,
Xcm = 0
⇒0=20+20×230⇒ 20+20×2=0⇒20=-20×2⇒x2=-1
Therefore, to keep the position of centre of mass unchanged, the block of mass 20 kg should be moved by a distance of 1 cm, towards left.
Answer: 9
Let the two masses m1 and m2 are kept along a vertical line.
It is given that:
m1 = 10 kg
m2 = 30 kg
First block is raised through a height of 7 cm.
Thus, to raise the centre of mass by 1 cm:
Ycm= m1y1 + m2y2m1 + m2⇒1= 10×7+30 × y240⇒70 + 30y2 = 40⇒30y2 = -30⇒y2 = -1
Therefore, the 30 kg block (m2) should be moved by 1 cm downwards in order to raise the centre of mass by 1 cm.
Answer: 10
As there is no gravity or other external forces acting on the system, the melting ice tends to acquire a spherical shape. Therefore, the centre of mass of the system does not move.
Answer: 11
Let the mass of the plate be M.
Consider a small semicircular portion of mass dm and radius r, as shown in fig.
dm = MπrdrπR22-R122 = MR22-R122rdr
The centre of mass is given as:
ycm=∫y dmM
ycm=∫R1R22rπ.MR22 – R122 × rdrM
=2πR22 – R122∫R1R2r2 dr =2πR22 – R122R23 – R133=4R12 + R1R2 + R223πR1 + R2
Answer: 12
It is given that:
Mass of Mr. Verma, m1 = 50 kg
Mass of Mr. Mathur, m2 = 60 kg
Mass of boat, m3 = 40 kg
Let A be the origin of the system (boat plus two men).
Initially, Mr. Verma and Mr. Mathur were at two extremes of the boat.
∴ Distance of the centre of mass:
Xcm=m1×x1 + m2×x2 + m3×x3m1 + m2 + m3=60×0 + 50×4 + 40×260 + 50 + 40=280150=1.87 m from A
As no external force is experienced in longitudinal direction, the centre of mass would not shift.
Initially, the centre of mass lies at a distance of 2 m from A.
When the men move towards the middle of the boat, the centre of mass shifts and lies at 1.87 m from A.
Therefore, the shift in centre of mass = 2 − 1.87 = 0.13 m, towards right
Hence, the boat moves 13 cm or 0.13 m towards right.
Answer: 13
The mass of the bob is m.
The mass of the cart is M.
Considering the bob falls at point A.
Initial distance of centre of mass of the system from P is given as
x=m×L+M×0M+m=mM+mL
When the bob falls in the slot, the distance of centre of mass of the system from P becomes zero.
∴ Shift in centre of mass = 0 – mLM + m=-mLM + m towards left=mLM + m towards right
Therefore, the cart moves a distance of mLM+m towards right.
Answer: 14
Given:
The mass of monkey is m.
The mass of balloon is M.
Initially, the monkey, balloon and the rope are at rest.
Let the centre of mass is at a point P.
When the monkey descends through a distance L,
The centre of mass shifts.
l0=m × L + M × 0M + m =mLM + m from P
Therefore, the balloon descends through a distance mLM + m.
Answer: 15
Let the masses of the two particles be m1 and m2.
Given:
m1 = 1 kg
m2 = 4 kg
Now,
Kinetic energy of the first particle = Kinetic energy of the second particle
12m1v12 = 12m2v22⇒m1m2 = v22v12⇒v2v1 = m1m2⇒v1v2 = m2m1
The ratio of linear momenta (mv) of the two particles,P1P2 = m1v1m2v2 = m1m2m2m1= m1m2 =14=12⇒P1:P2 = 1:2
Therefore, the ratio of linear momenta is 1:2.
Answer: 16
According to the question, uranium 238 nucleus emits an alpha-particle with a speed of 1.4 × 107m/s.
Let the speed of the residual nucleus thorium 234 be v2.
By the law of conservation of linear momentum, we have:
m1v1=m2v2
Here, m1 and v1 are the mass and velocity of the alpha-particle repectively, and m2 is the mass of the residual nucleus.
⇒4 × 1.4 × 107 = 234 × v2v2 = 4 × 1.4 × 107234 = 2.4 × 105 m/s
Therefore, the speed of the residual nucleus is 2.4×105 m/s.
Answer: 17
By the law of conservation of linear momentum, we have:
m1v1=m2v2
Here, m1 and v1 are the mass and velocity of the man respectively, and m2 and v2 are
the mass and velocity of the Earth respectively.
⇒ 50 × 1.8 = 6 × 1024×v2∴ v2 = 50 × 1.86 × 1024 = 15 × 10-24m/s⇒ v2 = 1.5 × 10-23m/s
Hence, the earth recoils with a speed of 1.5 × 10−23 m/s.
Answer: 18
It is given that:
Mass of proton, mp = 1.67 × 10−27 kg
Momentum of electron = 1.4 × 10−26 kg m/s
Momentum of antineutrino = 6.4 × 10−27 kg m/s
Let the recoil speed of the proton be Vp.
(a) When the electron and the antineutrino are ejected in the same direction,
Applying the law of conservation of momentum, we get:
mpVp = pelectron+pantineutrin o1.67 × 10-27×VP = 1.4×10-26 + 6.4×10-27 ⇒VP = 20.41.67 = 12.2 m/s, in the opposite direction
(b) When the electron and the antineutrino are ejected perpendicular to each other,
Total momentum of electron and antineutrino is given as,
p = pelectron + pantineutrino = (14)2 + (6.4)2 × 10-27 = 15.4 × 10-27 kg m/s⇒Vp = 15.4 × 10-27 kgm/s 1.67×10-27 kg⇒Vp = 9.2 m/s
Exercise : Solution of Questions on page Number : 161
Answer: 19
Mass of man = M
Initial velocity of the man = 0
Mass of bag = m
Let the man throws the bag towards left with a velocity v and himself moves towards right with a velocity V.
Answer: 20
It is given that:
Mass of the ball = 50 g = 0.05 Kg
Speed of the ball, v = 2.0 m/s
Incident angle = 45˚
v = 2 cos 45° i^-2 sin 45° j^v’ =-2 cos 45° i^-2 sin 45° j^(a) Change in momentum= mv→-mv→’ = 0.052 cos 45°i^-2 sin 45° j^-0.05-2 cos 45°i^-2 sin 45° j^= 0.1 cos 45° i^-0.1 sin 45° j^+0.1 cos 45°i^+0.1 sin 45° j^ = 0.2 cos 45° i^
∴ Magnitude=0.22= 0.14 Kg m/s
(b) The change in magnitude of the momentum of the ball,
P→2-P→1=2×0.5-2×0.5=0
i.e. There is no change in magnitude of the momentum of the ball.
Answer: 21
It is given that:
Wavelength of light = λ
Momentum of each photon = h/λ
Angle of incidence = θ
Answer: 22
As the block is exploded only because of its internal energy, the net external force on the system is zero.
Thus, the centre of mass of does not change.
Let the body was at the origin of the co-ordinate system during explosion.
Resultant velocity of two bodies of equal mass moving at a speed of 10 m/s in + x-axis and + y-axis direction, is given as:
If the centre of mass is at rest, the third part having equal mass as that of the other two masses will move in the opposite direction (i.e. 135° w.r.t. +x-axis) at the same velocity of 102 m/s..
Answer: 23
According to the question, the spaceship is removed from all other material objects and kept totally isolated from the surroundings. Thus, the mass loss by astronauts couldn’t slip away from the spaceship. Therefore, the total mass of the spaceship remains unchanged and so does its velocity.
Hence, the spaceship moves with the speed of 15 km/s.
Answer: 24
It is given that:
Diameter of hailstone = 1 cm = 0.01 m
⇒ Radius of hailstone, r = 0.005 m
Average speed of hailstone = 20 m/s
Density of hailstone = 900 kg/m3 = 0.9 g/cm3
Volume of the hailstones is given as,V = 43πr3 ⇒V = 43π(0.005)3 = 5.235 × 10-7 m3Mass = volume × density = 5.235 × 10-7× 900= 4.711×10-4 kg
∴ Mass of 2000 hailstone = 2000 × 4.711 × 10-4= 0.9422
Rate of change of momentum = 0.9422 × 20 ≈ 19 N/m2
∴ The total force exerted on the roof = 19×100=1900 N
Answer: 25
It is given that the mass of the ball is m.
Let the ball be dropped from a height h.
The speed of ball before the collision is v1.
Answer: 26
Given:
The mass of the railroad car is M.
The mass of the man is m.
The car recoils with a speed v, backwards on the rails.
Let the man of mass m approaches towards the engine with a velocity v’ w.r.t the engine.
∴ The velocity of man w.r.t earth is v’ − v, towards right.
Answer: 27
It is given that:
Mass of the car, the gun, the shells and the operator = 50m
Mass of one shell = m
Muzzle velocity of the shells, v = 200 m/s
Let the speed of car be v.
On applying the law of conservation of linear momentum, we get:
49mV+mv=0⇒ 49m × V + m ×200 = 0⇒ V= – 20049m/s (‘-‘ sign indicates direction towards left)
Thus, when another shell is fired, the velocity of the car with respect to the platform is,
V= 20049m/s, towards left
When one more shell is fired, the velocity of the car with respect to the platform is,
V1 = 20048m/s, towards left
∴ Velocity of the car w.r.t the earth = 200149+148 m/s
Answer: 28
It is given that:
Mass of each persons = m
Mass of railroad car = M
Let the velocity of the railroad w.r.t. earth, when the man on the left jumps off be V.
By the law of conservation of momentum:
Answer: 29
Given:
The mass of the small block is m.
Initial speed of this block is v.
The mass of the bigger block is M.
Initial speed of this block is zero.
At point A, as the small block transfers its momentum to the bigger block, both the blocks move with a common velocity V (say) in the same direction as v.
Using the law of conservation of linear momentum, we can write:
Initial momentum = final momentum
mv + M × O = (m + M)V
⇒ V = mvm + M
Therefore, the speed of the bigger block when the smaller block reaches point A of the surface ismvm+M
Exercise : Solution of Questions on page Number : 162
Answer: 30
It is given that:
Mass of the bugghi, mb = 200 kg
Velocity of the bugghi, Vb = 10 km/h
Mass of the boy, mboy = 25 kg
Velocity of the boy, VBoy = 4 km/h
Consider the boy and the bugghi as a system.
The total momentum before the process of sitting remains same after the process of sitting.
Using the law of conservation of momentum, we can write:
mbVb + mboy Vboy =(mb + mboy)V⇒200×10 + 25×4 = (200+25) × V⇒ V = 2100225 =283 Km/h
Answer: 31
It is given that:
Speed of the ball, v1 = 5.0 m/s
Mass of the ball, m1 = 0.5 kg
Mass of another ball, m2 = 1 kg
Let the velocity of this ball be v2 m/s.
On applying the law of conservation of momentum, we get:
m1v1 + m2v2 = 00.5×5 + 1×v2 = 0⇒ v2 = -2.5 m/s
Hence, the velocity of second ball is 2.5 m/s, opposite to the direction of motion of the first ball.
Answer: 32
It is given that:
Mass of the skater who is skating, m1 = 60 kg
Initial speed of this man, v1 = 10 m/s
Mass of the skater at rest, m2 = 40 kg
Initial speedof this man, v2 = 0
Let the velocity of both men after collision be v.
Using the law of conservation of momentum, we can write:
m1v1 + m2v2=(m1+m2)v⇒60 × 10 +0= 100 × v⇒v = 6 m/s The loss in kinetic energy during collision is given as,∆KE = 12m1v12 – 12(m1 + m2)v2⇒∆KE =12× 60 × (10)2 -12 × 100 × 36⇒∆KE = 1200 J
Answer: 33
It is given that:
Speed of the first particle during collision, v(t)=u1+t∆t(v1-u1)
Let v’ be the speed of the second particle, during collision.
On applying the law of conservation of linear momentum on both particles, we get:
m1u1 + m2u2 = m1v(t) + m2v‘
⇒m1u1 + m2u2 = m1u1 + m1×t∆t(v1-u1) + m2v’On dividing both the sides by m2, we get:u2 = m1m2t∆t(v1-u1) + v’⇒v’ = u2 – m1m2t∆t(v1-u1)
The speed of the second particle during collision can be written as a function of time and is given by the expression, u2-m1m2t∆t(v1-u1).
Answer: 34
Given:
The mass of bullet moving with speed v is m.
The mass of the ball is M and it is at rest.
m’ is the fractional mass of the ball that sticks with the bullet.
The remaining mass of the ball moves with the velocity v1.
Let v2 be the final velocity of the bullet plus fractional mass system.
On applying the law of conservation of momentum, we get:
mv + 0 = (m’ + m)v2 + (M − m‘) v1
⇒ v2 = mv- (M-m’)v1m + m’
Therefore, the velocity of the bullet after the collision is mv-(M-m’)v1m+m’ .
Answer: 35
Given:
The mass of the both balls is m.
Initial speed of first ball = v
Initial speed of second ball = 0
Let the final velocities of balls be v1 and v2 respectively.
e=velocity of separation velocity of approach⇒e=v1- v2v⇒v1- v2=ev …1
On applying the law of conservation of linear momentum, we get:
m(v1+v2)=mv ⇒v1+v2=v …(2)
According to the given condition, Final K.E.=34
Initial K.E.⇒12mv12 + 12mv22 = 34×12mv2⇒ v12 + v22 = 34v2⇒(v1+v2)2 + (v1-v2)22 = 34v2⇒1+e2v22 = 34v2
using the equations 1 and 2⇒ 1+e2 = 32⇒ e2 = 12⇒ e = 12
Hence, the coefficient of restitution is found to be 12.
Answer: 36
It is given that:
Mass of first block, m1 = 2 kg
Initial speed,v1 = 2.0 m/s
Mass of second block, m2 = 2 kg
Initial speed of this block = 0
For maximum possible loss in kinetic energy,
we assume that the collision is elastic and both the blocks move with same final velocity v (say).
On applying the law of conservation of linear momentum, we get:
m1v1 + m2 ×0 = (m1+m2)v
2 × 2 = (2 + 2)v
⇒ v = 1 m/s
Loss in K.E. in elastic collision is give by,
= 12m1v12 – 12m1+m2v= 12 × 2 × 22 – 12(2 + 2) × (1)2= 4 – 2 = 2 J
(b) The actual loss in K.E. = Maximum loss in K.E.2 = 1 J
Let the final velocities of the blocks be v1 and v2 respectively.
The coefficient of restitution is e.
∴ The loss in K.E. is given by,
12 × 2 × (2)2 – 122 × v12 – 12 × 2v22 = 1⇒4 – v12 + v22 = 1
⇒4 – 1 + e2 × 42 = 1⇒21 + e2 = 3⇒1 + e2 = 32⇒e2 = 12⇒e = 12
Answer: 37
It is given that:
Mass of particles = 100 g
Initial speed of the first particle = u
Final K.E. of the system after collision = 0.2J
Initial K.E. of the system, before collision = 12mu2 + 0
i.e. Initial K.E. = 12×0.1×u2 = 0.05 u2
Let v1 and v2 be the final velocities of the first and second block respectively.
By law of conservation of momentum, we know:
mv1+mv2=mu
⇒ v1+v2=u …(1)(v1-v2) + e(u1-u2) = 0⇒eu = v2-v1 …(2) [Putting u2=0,u1=u]Adding the equations (1) and (2),
we get : 2v2 = (1+e)u⇒v2 = u2(1+e)∴ v1 = u-u2(1+e) v1 = u2(1-e) According to given condition, 12mv12 + 12mv22 = 0.2⇒ v12 + v22 = 4⇒ u221+e2 = 4⇒ u2 = 81+e2
For maximum value of u, denominator should be minimum in the above equation.
i.e. e = 0
⇒ u2 = 8
⇒ u=22 m/s
For minimum value of u, denominator should have maximum value.
i.e. e = 1
⇒ u2 = 4
⇒ u = 2 m/s
Answer: 38
It is given that:
Weight of A = Weight of B = 40 kg
Velocity of ball = 5 m/s
(a) Case-1: Total momentum of the man A and ball remains constant.
∴ 0 = 4 × 5 − 40 × v
⇒ v = 0.5 m/s, towards left
(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.
⇒ 4 × 5 = 44 v
⇒ v=2044m/s
Case-3: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
⇒ 44 × 2044 = -4×5 + 40×v⇒ v = 1 m/s, (towards right)
Case-4 : When A catches the ball,
Applying the law of conservation of liner momentum, we get:
-4×5 + (-0.5)×40 = 44v⇒ v = 4044 = 1011 m/s, towards left
(c) Case-5: When A throws the ball,
Applying the law of conservation of linear momentum, we get:
44 × 1011 = 4×5 + 40×v⇒ v = 6040 = 32 m/s (towards left)
Case-6: When B receives the ball,
Applying the law of conservation of linear momentum, we get:
40×1 + 4×5 = 44×v⇒ v = 6044 m/s, towards right
Case-7: When B throws the ball,
On applying the law of conservation of linear momentum, we get:
⇒ v = 44 × 6044 m/s, towards right
Case-8: When A catches the ball,
On applying the law of conservation of linear momentum, we get:
-4×5 + 4032 = -44v⇒ V = -4044 = 1011 m/s, towards left
Similarly, after 5 round trips,
The velocity of A will be 5011 m/s and the velocity of B will be 5 m/s.
(d) As after 6 round trips, the velocity of A becomes 6011 i.e. > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.
(e) Let the ball and the body A be at origin, in the initial position.
∴ Xc = 40×0 + 4×0 + 40×d40 + 40 + 4 =1021d
Answer: 39
Let the velocity of the ball falling from height h1 be u (when it approaches the ground).
Velocity on the ground, u=2gh1
⇒u=2×9.8×2
Let the velocity of ball when it separates from the ground be v.(Assuming it goes up to height h2)
⇒ v= 2gh2= 2×9.8×1.5
Let the coefficient of restitution be e.
We know, v = eu
⇒e=2×9.8×1.52×9.8×2=32
Hence, the coefficient of restitution is 32.
Answer: 40
Let the nucleus recoils with a velocity v.
Applying the law of conservation of linear momentum, we get:
Linear momentum of recoiled nucleus = Linear momentum of gamma photon
⇒ mv = Ec
∴ v = Emc
Kinetic energy of the recoiled nucleus = 12Mv2
⇒K.E.=12mEmc2=12E2mc2
Decrease in the internal energy = photon energy + the kinetic energy of the recoiled nucleus
⇒ Decrease in the internal energy = E+E22mc2
Answer: 41
Given,
Mass of each block, MA = MB = 2 kg
Initial velocities of block A, Va = 1 m/s
Initial velocity of block B, Vb = 0
Spring constant of the spring = 100 N/m
Block A strikes the spring with a velocity of 1 m/s.
After the collision, it’s velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity V (say).
Using the law of conservation of energy, we get:
12MAVA2 + 12MBVB2 =12MAV2 + 12MBV2 + 12kx212 × 2(1)2 + 0 = 12 + 12 × v2 + 12×2 × 100
(where x is the maximum compression of the spring)
⇒ 1 − 2v2 = 50x2 …(1)
As there is no external force acting in the horizontal direction, the momentum is conserved.
⇒ MAVA + MBVB = (MA+MB)V⇒ 2×1 = 4×V⇒ V = 12m/s …(2)
Substituting this value of V in equation (1),
we get : 1 = 2×14 + 50×2⇒ 14 = 50×2
⇒ x2 = 1100 ⇒ x = 110m⇒x = 10 cm
Answer: 42
It is given that:
Mass of bullet, m = 20 g =0.02 kg
The initial speed, v1 = 500 m/s
Mass of block, M = 10 kg
The initial speed of block = 0
Final velocity of bullet, v2= 100 m/s
Let the final velocity of block when the bullet emerges out = v‘
Applying conservation of linear momentum,
mv1 + M × 0 = mv2 + Mv‘
⇒ 0.02 × 500 = 0.02 × 100 + 10 × v‘
⇒ v‘ = 0.8 m/s
Distance covered by the block, d = 20 cm = 0.02 m .
Let friction coefficient between the block and the surface = μ
Thus, the value of friction force, F=μmgF=μmg
Change in K.E. of block = Work done by the friction force
⇒12×M×0-12×M×v’2=μmgd⇒ 0-12×10×(0.8)2=μ×10×10×0.2⇒ μ=0.16
Answer: 43
Given:
Initial velocity of the projectile = u
Angle of projection of the projectile with respect to ground = θ
When the projectile hits the ground for the first time, the velocity remains same i.e. u.
The component of velocity parallel to ground, u cos θ should remain constant.
However, the vertical component of the projectile undergoes a change after the collision.
If the coefficient of restitution of collision between the projectile and the field is e,
The velocity of separation is given by,
⇒ v = eu sin θ
Therefore, for the second projectile motion,
Velocity of projection (u’) will be,
u’=(u cos θ)2+(eu sin θ)2Angle of projection, α = tan-1eu sin θu cos θ⇒α = tan-1(e tan θ)or, tan α = e tan θ …(2)Also, y=x tan α – gx2 sec2 α2u’2 …(3)Here, y=0
∴ tan α = e tan θ⇒sec2 α=1+e2 tan2 θand u’2=u2 cos2 θ+e2u2 sin2 θ
Putting the above calculated values in equation (3), we get:
xe tanθ = gx2(1+e2 tan2θ)2u2(cos2θ+e2 sin2θ)or, x =2eu2 tanθ (cos2θ+e2 sin2θ)g(1+e2 tan2θ)⇒x =2eu2 tanθ.cos2θg⇒x =eu2 sin2θg
Thus, from the starting point the projectile makes its second collision with the field at a distance,
x’=u2 sin 2θg+eu2 sin 2θg⇒x’=u2 sin 2θg(1+e)
Answer: 44
Given:
The angle of inclination of the inclined plane is θ.
A ball falls on the inclined plane from height h.
The coefficient of restitution is e.
Let the ball strikes the inclined plane at origin with velocity v0=2gh.
As the ball elastically rebounds, it recalls with the same velocity v0 at the same angle θ from the normal or y-axis.
Let the ball strikes the incline second time at any point P, which is at a distance l from the origin along the incline.
From the equation,
y = voyt + 12ayt2 we may write : 0 = v0 cos θ t – 12g cos θ t2
where t is time of motion of the ball in air, as the ball moves from origin to point P.
As t≠ 0, the value of t is 2v0g.
Now from the equation,
x=v0t + 12axt2we can write,l = v0 sin θ t + 12g sin θ t2
∴ l = v0 sin θ2v0g + 12g sin θ 2v0g2⇒l = 4v02 sin θg
Substituting the value of v0 in the above equation, we get:
l = 8h sin θ
Therefore, the ball again hits the plain at a distance, l = 8h sin θ
Answer: 45
Given:
The angle of inclination of the plane, θ = 45°
A ball falls on the inclined plane from a height (h) of 5 m.
The coefficient of restitution, e=34
The velocity with which ball strikes the inclined plane is given as,
v=2g×5=10 m/s
The ball makes an angle β with the horizontal, after the collision.
The horizontal component of velocity, 10cos45° remains unchanged.
However, the velocity in perpendicular direction to the plane after the collision will now be:
v1 = e×10 sin45°=34×10×12 =(3.75) 2 m/s
Similarly, v2 = 52 m/s Now,u = v22+v12 =50+28.125 =78.125 =8.83 m/s
Angle of reflection from the wall is given as,
β=tan-1(3.752)52=tan-134=37°
Angle of projection α = 90 − (θ + β)
⇒α = 90° − (45° + 37°) = 8°
Let the distance where the ball falls after the collision be L.
⇒ x = L cos θ
Angle of projection (α) = −8°
Now,
Y = x tan α – gx2 sec2 α2u2
⇒ −L sin θ = L cos θ × tan 8° – g2L2 cos2 θ sec2 80°(u)2
⇒ -sin 45° = cos 45°× tan 8° – 10 cos2 45 sec2 8°(8.83)2
On solving the above equation, we get:
L = 18.5 m
Answer: 46
It is given that:
Mass of block, M = 200 g = 0.20 kg
Mass of the particle, m = 120 gm = 0.12 kg
Height of the particle, h = 45 cm = 0.45 m
According to question, as the block attains equilibrium, the spring is stretched by a distance, x = 1.00 cm = 0.01 m.
i.e. M × g = K × x
⇒ 0.2 × g = K × x
⇒ 2 = K × 0.01
⇒ K = 200 N/m
The velocity with which the particle m strikes M is given by,
u = 2ghu = 2×10×0.45 = 9=3 m/s
After the collision, let the velocity of the particle and the block be V.
According to law of conservation of momentum, we can write:
mu = (m + M)V
Solving for V , we get:
V=0.12×30.32=98m/s
Let the spring be stretched through an extra deflection of δ.
On applying the law of conservation of energy, we can write:
Initial energy of the system before collision = Final energy of the system
⇒ 12mu2 + 12Kx2 = 12m+MV2 + 12Kx+δ2
Substituting appropriate values in the above equation, we get:
12×0.12×9 + 12×200×(0.01)2 = 120.32×8164 + 12×200×(δ+0.1)2
On solving the above equation, we get:
δ = 0.061
m = 6.1 cm
Exercise : Solution of Questions on page Number : 163
Answer: 47
Given:
Mass of bullet, m = 25 g = 0.025 kg
Mass of ballistic pendulum, M = 5 kg
Vertical displacement, h = 10 cm = 0.1 m
Let the bullet strikes the pendulum with a velocity u.
Let the final velocity be v.
Using the law of conservation of linear momentum, we can write:
mu = (M+m)v⇒ v = m(M+m)u⇒v = 0.255.025 × u = u201
Applying the law of conservation of energy, we get:
12(M+m)v2 = (M+m)gh⇒ u2(201)2 = 2×10×0.1⇒ u = 201×2 = 280 m/s
The bullet strikes the pendulum with a velocity of 280 m/s.
Answer: 48
Given:
Mass of bullet, m = 20 gm = 0.02 kg
Horizontal speed of the bullet, u = 300 m/s
Mass of wooden block, M = 500 gm = 0.5 kg
Let the bullet emerges out with velocity v.
Let the velocity of the block be v’.
Using the law of conservation of momentum, we get:
mu = Mv‘ + mv …(1)
Now, applying the work-energy principle for the block after the collision, we get:
0 – 12M×v’2 = -Mgh⇒ (v’)2 = 2gh v’=2gh =20×10×0.2 = 2 m/s
On substituting the value of v’ in equation (1), we get:
0.02×300 = 0.5×2 + 0.02×v⇒ v = 6-10.02 = 50.02⇒ v = 250 m/s
Hence, the speed of the bullet as it emerges out from the block is 250 m/s.
Answer: 49
It is given that two blocks of masses m1 and m2 are connected with a spring having spring constant k.
Initially the spring is stretched by a distance x0.
For the block to come to rest again,
Let the distance travelled by m1 be x1 (towards right), and that travelled by m2 be x2 towards left.
As no external force acts in horizontal direction, we can write:
m1x1 = m2x2 …(1)
As the energy is conserved in the spring, we get:
12kx02 = 12k(x1+x2-x0)2⇒ x0 = x1+x2-x0⇒ x1+x2 = 2×0 … (2)
∴ x1 = 2×0-x2Putting this value in equation 1, we get:m1(2×0-x0) = m2x2⇒2m1x0 – m1x2 = m2x2⇒x2 = 2m2m1+m2x0Similarly, x1 = 2m2m1+m2x0
Answer: 50
Given,
Velocity of mass, m2 = v0
Velocity of mass, m1 = 0
(a) Velocity of centre of mass is given by,
vcm = m1v1 + m2v2m1 + m2
⇒vcm = m1×0 + m2×v0m1 + m2⇒vcm =m2v0m1 + m2
(b) Let the maximum elongation in spring be x.
The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.
i.e. v1 = v2 = vcm
On applying the law of conservation of energy, we can write:
Change in kinetic energy = Potential energy stored in spring
⇒12m2v02 – 12(m1+m2)m2v0m1+m22 = 12kx2⇒ m2v02 1-m2m1+m2 = kx2
⇒x = v0m1m2m1+m2k1/2
Answer: 51
It is given that the force on both the blocks is F.
Let x1 and x2 be the extensions of blocks m1 and m2 respectively.
Total work done by the forces on the blocks = Fx1 + Fx2 …(1)
∴ Increase in the potential energy of spring = 12K(x1+x2)2 …(2)
Equating the equations (1) and (2), we get:⇒(x1+x2) = 2Fk …(3)
As the net external force on the system is zero, the centre of mass does not shift.
∴ m1x1 = m2x2 …(4)Using the equations 3 and 4,
we get : m1m2x1 + x1 = 2Fk⇒ x11+m1m2 = 2Fk⇒ x1 = 2Fm2k(m1+m2)
Similarly,
x2=2Fm1k(m1+m2)
Answer: 52
Given:
Force on block of mass, m1 = F1
Force on block of mass, m2 = F2
Let the acceleration produced in mass m1 be a1.
a1 = F1-F2m1+m2
Let the acceleration of mass m2 be a2.
a2=F2-F1m1+m2
Due to the force F2, the mass m1 experiences a pseudo force.
∴ Net force on m2=F2+m2a1
F” = F2 + m2×(F1 – F2)m1 + m2 = m1F2 + m2F2 + m2F1 – m2F2m1 + m2 = m1F2 + m2F1m1 + m2
Let m1 be displaced by a distance x1 and m2 be displaced by a distance x2.
Therefore, the maximum elongation of the spring = x1 + x2
Work done by the blocks = Energy stored in the spring
⇒ m2F1+m1F2m1+m2 × x1 × m2F1+m1F2m1+m2 × x2 = 12k(x1+x2)2⇒ x1+x2 = 2km1F2+m2F1m1+m2
Answer: 53
Given,
Mass of the man, Mm = 50 kg
Mass of the pillow, MP = 5 kg
Velocity of pillow w.r.t. man, V→pm = 8 ft/s
Let the velocity of man be V→m.
V→pm = V→p – (-V→m) = V→p + V→m⇒ V→p = V→pm – V→m
As no external force acts on the system, the acceleration of centre of mass is zero and the velocity of centre of mass remains constant.
Using the law of conservation of linear momentum:
Mm × Vm = Mp × Vp …(1)
Mm×Vm = Mp×(Vpm-Vm)⇒50×Vm = 5×(8-Vm)⇒Vm = 811 = 0.727 ft/s
∴ Velocity of pillow, V→p = 8 − 0.727 = 7.2 ft/s
The time taken to reach the floor is given by,
t=sv=87.2=1.1 s
As the mass of wall is much greater than the mass of the pillow,
velocity of block before the collision = velocity after the collision
⇒ time of ascent = 1.11 s
Hence, total time taken = 1.11 + 1.11 = 2.22 s
Answer: 54
Given:
Mass of the block, A = m
Mass of the block, B = 2m
Let the initial velocity of block A be u1 and the final velocity of block A,when it reaches the block B be v1.
Using the work-energy theorem for block A, we can write:
Gain in kinetic energy = Loss in potential energy
∴12mv12 – 12mu12 = mgh⇒v12 – u12 = 2gh⇒v1 = 2gh+u12 …(1)
Let the block B just manages to reach the man’s head.
i.e. the velocity of block B is zero at that point.
Again, applying the work-energy theorem for block B,
we get: 12×2m×(0)2 -12×2m×v2 = mgh⇒ v=2gh
Therefore, before the collision:Velocity of A, uA=v1 Velocity of B, uB=0
After the collision: Velocity of A, vA=v (say)Velocity of B, vB=2gh
As the collision is elastic, K.E. and momentum are conserved.
mv1+2m×0=mv+2m2gh⇒v1-v=22gh ………………..(2)
⇒ 12mv12 + 122m×(0)2 = 12mv2+122m2gh2⇒v12-v2 = 2×2gh×2gh ……..(3)
Dividing equation (3) by equation (2),
we get: v1+v = 2gh …………….(4)
Adding the equations 4 and 2,
we get: 2v1 = 32gh
Now using equation 1 to substitue the value of v1 ,
we get: 2gh+u2 = 322gh⇒ 2gh+u2 = 94(2gh)⇒ u = 2.5 gh
Block a should be started with a minimum velocity of 2.5 gh to get the sleeping man awakened.
Answer: 55
Given:
Mass of block = 490 gm
Initial speed of the block = 0
Mass of bullet = 10 gm
Initial speed of the bullet, v1 = 507 m/s
As the bullet gets embedded inside the block, this is an example of a perfectly inelastic collision.
Let the final velocity of the system (block and bullet) be VA.
Using the law of conservation of linear momentum, we get:
m1v1 + m2 × 0 = (m1 + m2)vA
⇒10 × 10 – 3 × 507 + 490 × 10 – 3 × 0 = (490 + 10) × 10 – 3 × vA
∴ vA=7 m/s
When the block loses contact at D, the component mg acts on it.
Let the velocity at D be vB.
m(vB)2r=mgsinθ(vB)2=gr sinθ …(1)
Using work energy theorem
12mvB2 – 12mvA2 =-mg(0.2 + 0.2sinθ)12grsinθ – 1272 =g(0.2 + 0.2sinθ)
3.5-12+9.8+0.2+sinθ=9.8+0.2 (1+sinθ)3.5-0.98 sinθ=1.96+1.96 sinθsinθ = 12⇒θ=30o
∴ Angle of projection = 90° − 30° = 60°
Time of reaching the ground,
⇒t=2hg=2×(0.2+0.2sin30o)9.8⇒t=0.247 s
Distance travelled in the horizontal direction is given by,
S = v cos θ × t
S = gr sin θ cos θ×t
S = 9.8 × 2 × 1232 × 0.247
S = 0.196 m
Total distance = (0.2 − 0.2 cos 30° + 0.196)
= 0.22 m
Answer: 56
Given:
Mass of the 1st ball = m
Mass of the 2nd ball = 2m
When the balls reach the lower end,
Let the velocity of m be v; and the velocity of 2m be v‘.
Using the work-energy theorem, we can write:
12×mv2-12×m(0)2 = mgl⇒ v = 2gl Similarly, velocity of block having mass 2m will be,
v’ = 2gl
∴ v=v’=u (say)
Let the velocities of m and 2m after the collision be v1 and v2 respectively.
Using the law of conservation of momentum, we can write:
m×u-2mu=mv1+2mv2⇒ v1+2v2 = -u …(1)On applying collision formula, we get: v1-v2 = [(u-v)]=-2u …(2)Substracting equation 1 from 2,
we get : 3v2 = u⇒ v2 = u3 = 2gl3
Substituting this value in equation (1),
we get : v1-v2 = -2u⇒ v1= -2u+v2 = -2u+u3 = -53u=-53×2gl = -50gl3
(b) Let the heights reached by balls 2m and m be h1 and h respectively.
Using the work energy principle, we get:
12×2m×(0)2 – 12×2m×v22 = -2m×g×h1⇒ h1=l9
Similarly,12×m×(0)2 – 12×mv12 = m×g×h2⇒12×50gl9 = g×h2⇒h2 = 25l9
As h2 is more than 2l, the velocity at the highest point will not be zero.
Also, the mass m will rise by a distance 2l.
Answer: 57
Given is a uniform chain of mass M and length L.
Let us consider a small element of chain at a distance x from the floor having length ‘dx‘.
Therefore, mass of element, dm=MLdx …(1)
The velocity with which the element strikes the floor is,
v=2gx …(2)
∴ The momentum transferred to the floor is,
P=(dm)v=ML. dx2gx
According to the given condition, the element comes to rest.
Thus, the force (say F1) exerted on the floor = Change in momentum
F1=dPdt=ML×dxdt2gx (for the chain element)As dxdt=v,F1 = ML×2gx2 using equation 2⇒F1 = ML2gx = 2MgxL
Again, the force exerted due to length x of the chain on the floor due to its own weight is given by,
W=ML(x)×g=MgxL
Thus, the total force exerted is given by.
F = F1 + W
where W is the weight of chain below the element dx up to the floor.
⇒ F=2MgxL+MgxL=3MgxL
Answer: 58
Given,
Speed of the block A = 10 m/s
The block B is kept at rest.
Coefficient of friction between floor and block B, μ = 0.10
Lets v1 and v2 be the velocities of A and B after collision respectively.
(a) If the collision is perfectly elastic, linear momentum is conserved.
Using the law of conservation of linear momentum, we can write:
mu1+mu2 = mv1+mv2⇒10+0=v1+v2 v1+v2=10 …1
We know,
Velocity of separation after collision =Velocity of approach
before collision v1-v2 = -(u1-v2)⇒v1-v2=-10 ….2
Substracting equation (2) from (1),
we get : 2v2 = 20⇒ v2=10 m/s
The deceleration of block B is calculated as follows:
Applying the work energy principle, we get:
12×m×(0)2-12×m×v2 = -m×a×s1⇒ -12×(10)2 = -μg×s1⇒s1 = 1002×1×10 = 50 m
(b) If the collision is perfectly inelastic, we can write:
m×u1+m×u2 = (m+m)×v⇒ m×10+m×0 = 2m×v⇒ v = 102 = 5m/s
The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:
12×2m×(0)2 – 12×2m×(v)2 = 2m×μg×s2⇒(5)20.1×10×2 = s2⇒ s2 = 12.5 m
Answer: 59
Given:
Initial velocity of 2 kg block, v1 = 1.0 m/s
Initial velocity of the 4 kg block, v2 = 0
Let the velocity of 2 kg block, just before the collision be u1.
Using the work-energy theorem on the block of 2 kg mass:
The separation between two blocks, s = 16 cm = 0.16 m
∴12m×u12-12m×(1)2 = -μ×mg×s ⇒ u1 = (1)2-2×0.20×10×0.16⇒ u1 = 0.6 m/s
As the collision is perfectly elastic, linear momentum is conserved.
Let v1, v2 be the velocities of 2 kg and 4 kg blocks, just after collision.
Using the law of conservation of linear momentum, we can write:
m1u1+m2u2=m1v1+m2v2⇒ 2×0.6+4×0=2v1+4v2⇒ 2v1+4v2=1.2 …(1)
For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)
i.e. v1-v2 = +(u1-u2)=+(0.6-0)⇒v1-v2 = -0.6 …(2)
Subtracting equation (2) from (1),
we get: 3v2 = 1.2⇒ v2 = 0.4 m/s
∴ v1 = -0.6+0.4 = -0.2 m/s
Let the 2 kg block covers a distance of S1.
∴ Applying work-energy theorem for this block, when it comes to rest:
12×2×(0)2 + 12×2×(0.2)2 = -2×0.2×10×S1⇒ S1 =1cm.
Let the 4 kg block covers a distance of S2.
Applying work energy principle for this block:
12×4×(0)2 – 12×4×(0.4)2 = -4×0.2×10×S2⇒ 2×0.4×0.4 = 4×0.2×10×S2⇒ S2 = 4cm
Therefore, the distance between the 2 kg and 4 kg block is given as,
S1 + S2 = 1 + 4 = 5 cm
Answer: 60
According to the question, the surface is frictionless. Thus, the block m will slide down the inclined plane of mass M.
Acceleration, a1 = g sin α (Relative to the inclined plane)
The horizontal component of acceleration a1 is given by ax = g sin α cos α, for which the block Maccelerates towards left.
Let the left acceleration be a2.
By the concept of centre of mass, we can say that the external force is zero in the horizontal direction.
max = (M+m)a2
Absolute (resultant) acceleration of m on the plane M, along the direction of the incline will be = a= g sin α- a2 cos α
Let the time taken by the block m to reach the bottom end be t.
Now,
s=ut+12at2⇒ hsin α=12at2⇒ t=2ha sin α
Thus, the velocity of the bigger block after time t will be,
vm=u=a2t=mg sinα cos α M+m 2ha sin α=2m2g2h sin2 α cos2 α(M+m)2 a sin α1/2
Subtracting the value of a from equation (2), we get:
vM= 2m2g2h sin2 α(M+m)2 sin α×cos2 αg sin α(M+m)(M+m sin2 α)
Answer: 61
,
According to the question, mass m is given with a speed v over the larger mass M.
(a) When the smaller block travels on the vertical part, let the velocity of the bigger block be v1, towards left.
From the law of conservation of momentum (in the horizontal direction), we get:
mv=(M+m)v1⇒ v1=mvM+m
(b) When the smaller block breaks off, let its resultant velocity be v2.
Using the law a of conservation of energy, we get:
12mv2 = 12Mv1 2+ 12mv22+mgh⇒v22 = v2-Mmv12-2gh …(i)
⇒v22 = v21-Mm.m2(M+m)2-2gh⇒ v2 = m2+Mm+m2v2(M+m)2-2gh1/2
(c) Vertical component of the velocity v2 of mass m is given by,
vy2=v22-v12= M2+Mm+m2M+m2v2 – 2gh – m2v2M+m2
∴ v1=mvM+m⇒vy2 = M2+Mm+m2-m2M+m2v2 – 2gh⇒vy2 = Mv2M+m-2gh …(ii)
To determine the maximum height (from the ground),
Let us assume that the body rises to a height h1 over and above h.
Now,
12mvy2 = mgh1⇒ h1 = vy22g …(iii)
∴Total height,
H=h+h1⇒H = h+vy22g⇒H = h+Mv2(M+m)2g-h
⇒H=Mv2(M+m)2g
(d) Because the smaller mass also has a horizontal component of velocity V1 at the time it breaks off from M (that has a velocity v1), the block m will again land on the block M.
The time of flight of block m after it breaks off is calculated as:
During the upward motion (BC),
0 = vy-gt⇒ t1 = vyg = 1gMv2(M+m)-2gh …(4)
Thus, the time for which the smaller block is in flight is given by,
T=2t1=2gMv2-2(M+m)gh(M+m)
The distance travelled by the bigger block during this time is,
S = v1T
Exercise : Solution of Questions on page Number : 164
Answer: 62
It is given that h is much lesser than the radius of the earth.
Mass of the earth, Me = 6 × 1024 kg
Mass of the block, Mb = 3 × 1024 kg
Let Ve be the velocity of earth and Vb be the velocity of the block.
Let the earth and the block be attracted by gravitational force.
Thus, according to the conservation law of energy, the change in the gravitational potential energy will be the K.E. of block.
GMeMb1R+h2-1R+h = 12Me×Ve2+12Mb×Vb2 …(1)
As only the internal force acts in this system, the momentum is conserved.
MeVe = MbVb
⇒Ve=MbVbMe …(ii)
Answer: 63
According to the question, the collision of the two bodies of mass m is not head-on. Thus, the two bodies move in different directions.
Let the velocity vectors of the two bodies after collision be v1 and v2.
As the collision in the question is elastic, momentum is conserved.
On applying the law of conservation of momentum in X-direction, we get:
mu1+m×0 = mv1 cos α+mv2 cos β …(i)
On applying the law of conservation of momentum in Y-direction, we get:
0=mv1 sin α -mv2 sin β …(ii)
⇒mv1 sin α =mv2 sin βNow,12mu12 + 0 = 12mv12 + 12m×v22
⇒ u12 = v12+v22….(iii)
On squaring equation (i),
we get:u12 = v12cos2 α + v22 cos2β + 2v1v2 cos α cos βEquating the equations (i) and (iii),
we get : v12+v22 = v12 cos2α + v22 cos2β + 2v1v2 cosα cosβ
⇒v12 sin2α + v22 sin2β = 2v1v2 cosα cosβ
⇒2v12 sin2α = 2×v1×v1 sinα sinβ cosα cosβ
⇒sinα sinβ = cosα cosβ
⇒ cosα . cosβ – sinα cosβ = 0
⇒ cos(α+β)= 0=cos 90°⇒ α+β=90°
Answer: 64
It is given that the mass of both the bodies (small particle and spherical body) is same.
Let the velocity of the particle be v.
We break the particle velocity into two components: v cosα (normal to the sphere) and v sinα (tangential to the sphere).
From figure,
cosα=r2-ρ2 and sinα =ρr
Here, the collision occurs due to the component v cosα.
After an elastic collision, bodies of same mass exchange their respective velocities.
Thus, the spherical body will have a velocity v cosα, while the particle will not have any component of velocity in this direction.
However, the tangential velocity of the particle v sinα will remain unaffected.
Thus, we have:
Velocity of the sphere = v cosα=vrr2-ρ2
And,
Velocity of the particle = v sinα = vρr