KKMishra's Tutorials: Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, OBJECTIVE-I

My Channel on YouTube → SimplePhysics with KK

For links to

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
OBJECTIVE-I

1. Consider the following two equations:

(A) R =(1/M)Σimiri and

(B) aCM = F/M

In a non-inertial frame

(a) both are correct

(b) both are wrong

(d) B is correct but A is wrong.

Answer: (c)

Explanation: When motions are analyzed in a non-inertial frame, a pseudo force equal to ma is applied to the body. Where m is mass and a is the acceleration of the frame. As we can see, with the application of this pseudo force R in (A) is not affected bur F becomes F-ma. So the acceleration of CoM is changed to (F-ma)/M. Hence (B) is not correct.

2. Consider the following statements:

(A) Linear momentum of the system remains constant.

(B) Center of mass of the system remains at rest.

(a) A implies B and B implies A.

(b) A does not imply B and B does not imply A

(d) B implies A but A does not imply B.

Answer: (d)

Explanation: Position of CoM is given by R =(1/M)Σimiri . If we differentiate both sides w.r.t. time t, we get,
dR/dt =(1/M)Σimi.dri/dt
vcm=(1/M)Σimivi
Now if (B) is true, it means vcm = 0, → Σimivi = 0 =Constant. So (B) implies (A). Now let us check if (A) is true but not equal to zero → Σimivi = k
Then vcm=(1/M)k =k/M =not equal to zero.

i.e. CoM of the system is not at rest. So (A) does not imply (B).

3. Consider the following two statements:

(A) Linear momentum of a system of particles is zero.

(B) Kinetic energy of a system of particles is zero.

(a) A implies B and B implies A.

(b) A does not imply B and B does not imply A

(d) B implies A but A does not imply B.

Answer: (d)

Explanation: If (B) is true, then ½Σimivi² = 0. In this equation v is magnitude of velocity and m is mass. Mass cannot be zero and square of a scaler quantity can only be zero if it is zero. It means magnitude of velocity of each particle is zero. In that case Σimivi =0. So clearly (B) implies (A).
Now if (A) is true, it does not mean that magnitude of each particle is zero. Since linear momentum is a vector, and sum (resultant) of vectors can be zero even if each vector is non-zero. It means momentum of each particle is not zero, hence some of the particles may have non-zero magnitude. In that case (B) is not true. So (A) does not imply (B).

4. Consider the following two statements:

(A) The linear momentum of a particle is independent of the frame of reference.

(B) The kinetic energy of a particle is independent of the frame of reference.

(a) Both A and B are true.

(b) A is true but B is false.

(d) both A and B are false.

Answer: (d)

Explanation: With the application of pseudo force in a noninertial frame of reference velocities change, so do linear momentum and kinetic energy because both depend on velocity. Hence both statements are false.

5. All the particles of a body are situated at a distance R from the origin. The distance of the center of mass of the body from the origin is

(a) = R

(b) ≤ R

(d) ≥ R

Answer: (b)

Explanation: Since all the particles are at a distance of R from the origin, it implies that the body is in a shape of an arc of a circle. Let the mass of the body be M and position vectors of its ends make angles θ1 and θ2. Mass per unit length = M/R(θ2-θ1). See figure below.

Figure for Q-5

Now for center of mass,
X=(1/M) ∫{M/R(θ2-θ1)}R.dθ*R.cosθ ...(Between limits θ2 and θ1)
= R/(θ2-θ1)∫ cosθdθ ........(Between limits θ2 and θ1)
= R/(θ2-θ1)[sinθ] ........(Between limits θ2 and θ1)
= R.(sinθ2-sinθ1)/(θ2-θ1)
And
Y=(1/M) ∫{M/R(θ2-θ1)}R.dθ*R.sinθ ...(Between limits θ2 and θ1)
= R/(θ2-θ1)∫ sinθdθ ........(Between limits θ2 and θ1)
= R/(θ2-θ1)[cosθ] ........(Between limits θ2 and θ1)
= R.(cosθ1-cosθ2)/(θ2-θ1)
Now Rcom = √(X²+Y²)
={R/(θ2-θ1)}√[(sinθ2-sinθ1)²+(cosθ1-cosθ2)²]
={R/(θ2-θ1)}*√[sin²θ2+sin²θ1-2sinθ1.sinθ2+cos²θ1+cos²θ2-2cosθ1cosθ2]

={R/(θ2-θ1)}*√[2-2(sinθ1.sinθ2+cosθ1cosθ2]

={R/(θ2-θ1)}*√[2-2cos(θ2-θ1)]

={R/(θ2-θ1)}*√[2*2sin²(θ2-θ1)/2]

={R/(θ2-θ1)}*2sin(θ2-θ1)/2

=R*[sin(θ2-θ1)/2]/[(θ2-θ1)/2]

=R*sinβ/β {Where β=(θ2-θ1)/2}

But sinβ/β≤1 Its limiting value is maximum = 1 when β→0. In this situation, θ2-θ1 =0, i.e. the arc in question reduces to a point mass. As β gradually increases above zero, sinβ/β gradually decreases from 1. So Rcom ≤ R.

Note: Consider the case when β=π (i.e. θ2-θ1=2π, which means the arc becomes circle)
Rcom = R*sinπ/π = 0. Which means the center of mass is at the origin or center of the circle.

6. A Circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure (9-Q2). The density of the material and the thickness are same everywhere.

The center of mass of the composite system will be

(a) inside the circular plate

(b) inside the square plate

(d) outside the system

Answer: (b)

Explanation: The mass of circular plate and the square can be assumed as concentrated at their centers. Hence CoM of the system will be on the line joining their centers. Since the mass of square plate will be more than the circular plate due to the greater area, hence combined CoM will be towards square plate from the middle i.e. inside the square plate.

7. Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration a. The center of mass has an acceleration

(a) zero

(b) ½a

(d) 2a

Answer: (b)

Explanation: Let the distance of CoM from the origin be X. if each particles with mass m be at distances x1 and x2 then
X= (m*x1+ m*x2)/(m+m)
Velocity of CoM = V =dX/dt = (v1+ v2)/2
Acceleration of CoM = dV/dt = (a1+ a2)/2
since a1 =0 and a2 =a,
Acceleration of CoM = dV/dt = (0+a)/2 =½a

8. Internal forces can change

(a) the linear momentum but not the kinetic energy

(b) the kinetic energy but not the linear momentum

(d) neither the linear momentum nor the kinetic energy.

Answer: (b)

Explanation: If there is no external force the linear momentum is conserved because the velocity of CoM is not changed, so internal forces cannot change the linear momentum of the system. Since internal forces can change the velocity of the constituent particles of the system, K.E. of the system can change because even for negative velocity K.E. =½mv² will be positive.

9. A bullet hits a block kept at rest on a smooth horizontal surface and gets embedded into it. Which of the following does not change.

(a) Linear momentum of the block

(b) kinetic energy of the block

(d) temperature of the block.

Answer: (c)

Explanation: In this case velocity of the block will change, hence (a) and (b) will change. Some part of the K.E. of the bullet will be converted into heat energy due to friction between bullet and the block during the embedment. Hence (d) will also change. Since height of the block does not change hence (c) will not change.

10. A uniform sphere is placed on a smooth horizontal surface and a horizontal force F is applied on it at a distance h above the surface. The acceleration of the center

(a) is maximum when h = 0

(b) is maximum when h = R

(d) is independent of h.

Answer: (d)

Explanation: Since the surface is smooth, no other force is there along the force and the sphere will slide with an acceleration =F/m which is independent of h.

11. A body falling vertically downwards under gravity breaks in two parts of unequal masses. The center of mass of the two parts taken together shifts horizontally towards

(a) heavier piece

(b) lighter piece

(d) depends on the vertical velocity at the time of breaking.

Answer: (c)

Explanation: Since the body breaks due to internal forces, the center of mass of all parts taken together will remain the same and will not shift horizontally.

12. A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the center of mass

(a) of the box remains constant

(b) of the box plus the ball system remains constant

(d) of the ball relative to the box remains constant.

Answer: (b)

Explanation: In each collision, the velocities of the ball and the box will change individually, hence (a), (b) and (d) are not correct. The system of ball and the box together will have its center of mass constant because the forces of collisions will be the internal forces then.

13. A body at rest break into two pieces of equal masses. The parts will move

(a) in same direction

(b) along different lines

(d) in opposite directions with unequal speeds.

Answer: (c)

Explanation: Let the velocities of two parts be v and v' and mass of each part = m. Momentum before the break = 2m*0 = 0.
Momentum after the break =mv+mv'. Since there is no external force on the body, the linear momentum will be conserved.
→mv+mv' = 0
→v = -v'
So the parts will move in opposite directions with equal speeds.

14. A heavy ring of mass m is clamped on the periphery of a light circular disc. A small particle having equal mass is clamped at the center of the disc. The system is rotated in such a way that the center moves in a circle of radius r with a uniform speed v. We conclude that an external force

(a) mv²/r must be acting on the central particle

(b) 2mv²/r must be acting on the central particle

(d) 2mv²/r must be acting on the ring.

Answer: (c)

Explanation: Since the heavy ring and the small particle have equal masses, the center of mass of the system will be at r/2 from the ring. Hence the force acting on the system =mv²/(r/2) = 2mv²/r.

15. The quantities remaining constant in a collision are

(a) momentum, kinetic energy and temperature

(b) momentum and kinetic energy but not temperature

(d) momentum, but neither kinetic energy nor temperature.

Answer: (d)

Explanation: Since velocities of the colliding bodies change, the K.E. will change. The shapes of the bodies at the time of collision change due to the force, hence temperature will also change. Only the momentum will remain constant.

16. A nucleus moving with a velocity v emits an ɑ-particle. Let the velocities of the ɑ-particle and the remaining nucleus be v1 and v2 and their masses be m1 and m2

(a) v, v1 and v2 must be parallel to each other.

(b) None of the two of v, v1 and v2 should be parallel to each other.

(d) m1v1 + m2v2 must be parallel to v.

Answer: (d)

Explanation: Linear momentum, which is a vector, is conserved of a system if no external force is acted upon. The direction of the velocity of the nucleus v is same as its momentum. The momentum after emission of ɑ-particle m1v1 + m2v2 will have the same direction as v due to conservation principle.

17. A shell is fired from a cannon with a velocity V at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is

(a) 3V.cosθ

(b) 2V.cosθ

(d) V.cosθ

Answer: (a)

Explanation: Component of the velocity in horizontal direction = V.cosθ. Hence the momentum before the explosion in the horizontal direction = 2m.Vcosθ. Since one part retraces its path to canon, so its velocity immediately after the explosion = -V.cosθ and let the speed of the other piece be V'. From the conservation principle of the momentum,
mV'-mV.cosθ = 2m.Vcosθ
→V' = 3Vcosθ

18. In an elastic collision

(a) the initial kinetic energy is equal to the final kinetic energy

(b) the final kinetic energy is less than the initial kinetic energy

(d) the kinetic energy first increases then decreases.

Answer: (a)

Explanation: The initial kinetic energy before the collision and the final kinetic energy after the collision will remain the same. We cannot say that (c) is true because during the contact a part of the kintic energy is stored as elastic potential energy in the bodies, and at this time total kinetic energy is less than the initial or final kinetic energy.

19. In an inelastic collision

(a) the initial kinetic energy is equal to the final kinetic energy

(b) the final kinetic energy is less than the initial kinetic energy

(d) the kinetic energy first increases then decreases.

Answer: (b)

Explanation: If the collision is perfectly inelastic the deformation of the bodies during the collision is permanent and the part of initial kinetic energy lost in deformation is not recovered. As a result, the final kinetic energy is less than the initial kinetic energy.
If the collision is partly inelastic, the deformations of the bodies are partly recovered and still a part of the initial kinetic energy remains utilized in the partial deformation. Hence here again the final kinetic energy is less than the initial kinetic energy. Though the difference, in this case, is less than the perfectly inelastic collision.

===<<<O>>>===