Q#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
(c) is maximum when h = 2R
(d) is independent of h.
Answer: (d)
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.
Q#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
(c) does not shift horizontally
(d) depends on the vertical velocity at the time of breaking.
Answer: (c)
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.
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
(c) of the ball remains constant
(d) of the ball relative to the box remains constant.
Answer: (b)
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.
A body at rest break into two pieces of equal masses. The parts will move
(a) in same direction
(b) along different lines
(c) in opposite directions with equal speeds
(d) in opposite directions with unequal speeds.
Answer: (c)
Let the velocities of two parts be v and v' and mass of each part = m. Momentum before the break = 2m x 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.
Q#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
(c) 2mv²/r must be acting on the system
(d) 2mv²/r must be acting on the ring.
Answer: (c)
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.
The quantities remaining constant in a collision are
(a) momentum, kinetic energy and temperature
(b) momentum and kinetic energy but not temperature
(c) momentum and temperature but not kinetic energy
(d) momentum, but neither kinetic energy nor temperature.
Answer: (d)
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.
Q#16
A nucleus moving with a velocity vemits an ɑ-particle. Let the velocities of the ɑ-particle and the remaining nucleus be v1and 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.
(c) v1 + v2 must be parallel to v.
(d) m1v1 + m2v2 must be parallel to v.
Answer: (d)
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.
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) 3Vcosθ
(b) 2Vcosθ
(c) (3/2) Vcosθ
(d) Vcosθ
Answer: (a)
Component of the velocity in horizontal direction = Vcosθ. Hence the momentum before the explosion in the horizontal direction = 2mVcosθ. Since one part retraces its path to canon, so its velocity immediately after the explosion = -Vcosθ and let the speed of the other piece be V'. From the conservation principle of the momentum,
mV' – mVcosθ = 2mVcosθ
V' = 3Vcosθ
Q#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
(c) the kinetic energy remains constant
(d) the kinetic energy first increases then decreases.
Answer: (a)
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.
Q#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
(c) the kinetic energy remains constant.
(d) the kinetic energy first increases then decreases.
Answer: (b)
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.
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