Q#1
Consider the following two equations:(A) R = (1/M)Σmiri and
(B) aCM = F/M
In a non-inertial frame
(a) both are correct
(b) both are wrong
(c) A is correct but B is wrong
(d) B is correct but A is wrong.
Answer: (c)
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.
Q#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
(c) A implies B but 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) Σmiri . If we differentiate both sides w.r.t. time t, we get,
dR/dt = (1/M) Σmidri/dt
vcm = (1/M) Σmivi
Now if (B) is true, it means vcm = 0, → Σmivi = 0 = Constant. So, (B) implies (A). Now let us check if (A) is true but not equal to zero → Σmivi = 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).
Q#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
(c) A implies B but B does not imply A
(d) B implies A but A does not imply B.
Answer: (d)
Explanation: If (B) is true, then ½Σmivi² = 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 Σmivi = 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).
Q#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.
(c) A is false but B is true.
(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.
Q#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
(c) > R
(d) ≥ R
Answer: (b)
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.
Now for center of mass,
X = (1/M) ∫{M/R(θ2 – θ1)}R.dθ(Rcosθ) ...(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 – cos θ2)/(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]
=Rsinβ/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, R com ≤ R.
Note: Consider the case when β=π (i.e. θ2 - θ1=2π, which means the arc becomes circle)
Rcom = Rsinπ/π = 0. Which means the center of mass is at the origin or center of the circle.
Q#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
(c) at the point of contact
(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.
Q#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
(c) a
(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 = (mx1+ mx2)/(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
Q#8
Internal forces can change
(a) the linear momentum but not the kinetic energy
(b) the kinetic energy but not the linear momentum
(c) linear momentum as well as kinetic energy
(d) neither the linear momentum nor the kinetic energy.
Answer: (b)
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.
Q#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
(c) gravitational potential energy of the block
(d) temperature of the block.
Answer: (c)
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.
X = (1/M) ∫{M/R(θ2 – θ1)}R.dθ(Rcosθ) ...(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 – cos θ2)/(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]
=Rsinβ/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, R com ≤ R.
Note: Consider the case when β=π (i.e. θ2 - θ1=2π, which means the arc becomes circle)
Rcom = Rsinπ/π = 0. Which means the center of mass is at the origin or center of the circle.
Q#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
(c) at the point of contact
(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.
Q#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
(c) a
(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 = (mx1+ mx2)/(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
Q#8
Internal forces can change
(a) the linear momentum but not the kinetic energy
(b) the kinetic energy but not the linear momentum
(c) linear momentum as well as kinetic energy
(d) neither the linear momentum nor the kinetic energy.
Answer: (b)
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.
Q#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
(c) gravitational potential energy of the block
(d) temperature of the block.
Answer: (c)
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.
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