Q#1
The center of mass of a system of particles is at the origin. It follows that(a) the number of particles to the right of the origin is equal to the number of particles to the left
(b) the total mass of the particles to the right of the origin is same as the total mass to the left of the origin
(c) The number of particles on X-axis should be equal to the number of particles on Y-axis
(d) if there is a particle on the positive X-axis, there must be at least one particle on the negative X-axis.
Answer: None
X = (1/M)Σmixi
Since CoM is at the origin, X = 0.
It means Σmixi = 0. But it does not imply any of the above.
Q#2
A body has its center of mass at the origin. The x-coordinate of the particles
(a) may be all positive
(b) may be all negative
(c) may be all non-negative
(d) may be positive for some cases and negative in other cases.
Answer: (c), (d)
As in the above explanation, for Σimixi = 0, the x-coordinate of the particles may be positive for some cases and negative in other cases as in the option (d).
If all the particles lie on the y-axis, then CoM may lie at the origin in some cases. In such situation x-coordinates of all the particles are non-negative. Option (c).
Q#3
In which of the following cases the center of mass of a rod is certainly not at its center?(a) The density continuously increases from left to right
(b) The density continuously decreases from left to right
(c) The density decreases from left to right up to the center and then increases
(d) The density increases from left to right up to the center and then decreases.
Answer: (a), (b)
Assume the parts of the rod on left and right from the middle as two separate bodies. Due to the continuous variation of the density, the masses of each body will be different. The CoM of the lighter body will be closer than that of the heavier body. If we take the middle of the rod as origin then CoM of the rod will be the CoM of these two bodies taking their masses at their CoM, Now Σmx for the heavier part will be greater than the lighter part. Hence total Σmx will not be zero and the CoM of the rod will not be in the middle.
Q#4
If the external forces acting on a system have zero resultant, the center of mass
(a) must not move
(b) must not accelerate
(c) may move
(d) may accelerate.
Answer: (b), (c)
If there is no/zero resultant force acting on a system, there will be no acceleration. Hence (b), not (d).
If it is already moving with a uniform velocity, it will continue its motion without acceleration. Hence (c), not (a).
Q#5
A nonzero external force acts on a system of particles. The velocity and the acceleration of the center of mass are found to be vo and ao at an instant t. It is possible that
(a) vo = 0, ao = 0
(b) vo = 0, ao ≠ 0
(c) vo ≠ 0, ao = 0
(d) vo ≠ 0, ao ≠ 0
Answer: (b), (d)
Since the external force is nonzero, there will be nonzero acceleration. Hence (b) and (d).
It may be at rest at time t, then (b) otherwise (d).
Q#6
Two balls are thrown simultaneously in the air. The acceleration of the center of mass of the two balls while in air
(a) depends on the direction of the motion of the balls
(b) depends on the masses of the two balls
(c) depends on the speed of the two balls
(d) is equal to g.
Answer: (d)
Since there is no force in the horizontal direction there will be no acceleration of CoM in the horizontal direction whatever be the direction, mass or speed of the two balls. Only force is the gravitational force hence acceleration of CoM will be acceleration due to gravity g.
Q#7
A block moving in air breaks into two parts and the parts separate
(a) the total momentum must be conserved
(b) the total kinetic energy must be conserved
(c) the total momentum must change
(d) the total kinetic energy must change.
Answer: (a), (d)
When the block moving in air breaks into two parts due to internal forces, the speeds of the parts will change. Since the Kinetic energy is proportional to the square of the speed it will be always positive. Hence the total kinetic energy must change.
The momentum is a vector and even if the velocities of the parts are different in magnitude and direction the vector addition of momentums will be same as before the split. Because in the absence of external force the total momentum is conserved.
Q#8
In an elastic collision
(a) the kinetic energy remains constant
(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.
Answer: (b), (c), (d)
In a fully elastic collision only during the contact period a part of the kinetic energy is stored as elastic potential energy in the bodies and their shapes are slightly changed. Since the collision is elastic the bodies release the stored elastic potential energy and regain their shape. Again the total energy is in the form of K.E. Hence the final K.E. is equal to the initial K.E. But we can not say that the K.E. remains constant because during the contact period it is reduced. So (c) is true and (a) false.
When there is no external force the momentum is conserved. Hence (b) and (d) is true.
Q#9
A ball hits a floor and rebounds after an inelastic collision. In this case
(a) the momentum of the ball just after the collision is same as that just before the collision
(b) the mechanical energy of the ball remains the same during the collision
(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.
Answer: (c), (d)
Since the velocity before and after the collision change hence momentum of the ball will change. (a) is not true.
Since the collision is inelastic a part of the mechanical energy is lost hence (b) is not true.
Taking earth and the ball as a system there is no external force on the system. Hence the total momentum of the ball and the earth is conserved. (c) is true.
From the conservation principle of the energy, the total energy of the ball and the earth remains the same. (d) is true.
Q#10
A body moving towards a finite body at rest collides with it. It is possible that
(a) both the bodies come to rest
(b) both the bodies move after collision
(c) the moving body comes to rest and the stationary body starts moving
(d) the stationary body remains stationary, the moving body changes its velocity.
Answer: (b), (c)
Since one of the bodies has some velocity before collision hence total momentum of the bodies is nonzero. Since in the absence of external force total momentum is conserved, both the bodies cannot come to rest. Because then the total momentum will become zero. (a) is not true.
In both the options (b) and (c) it is possible that final momentum is equal to the initial momentum.
The option (d) is not possible because the total momentum will not remain the same as before the collision.
Q#11
In head-on elastic collision of two bodies of equal masses
(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.
Answer: all
If u and v are the velocities before collision and u' and v' are the velocities after collision, then we have
u' = (m – m)u/(m+m) + 2mv/(m + m) = 0 + v = v
and v' = 2mu/(m + m) + (m – m)v/(m + m) = u + 0 = u
So the velocities and speeds are interchanged. Hence (a) and (b) are true.
Since the velocities are interchanged and masses are equal hence the momenta are also interchanged. Hence (c) is true.
If u > v then after the collision the speeds of bodies are interchanged. Now the faster body slows down and the slower body speeds up. Hence (d) is true.
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