As you know, momentum and kinetic energy are conserved in elastic collisions. But kinetic energy is not conserved in inelastic collisions, eventhough momentum is conserved. Consider the following M.C.Q.
A steel sphere of mass 20 gram moving horizontally with a velocity of 2m/s collides elastically with the bob of a long simple pendulum at rest. If the mass of the pendulum bob is 20 gram what is the height up to which the bob is raised?
(a) 10cm (b) 20cm (c) 40cm (d) 50cm (e) 60cm
As the collision is elastic and the masses of the sphere and the bob are the same, the sphere will transfer the entire kinetic energy to the bob. (This can be easily proved by equating the total initial momentum and the total initial kinetic energy to the total final momentum and the total final kinetic energy respectively). After receiving the kinetic energy from the sphere, the pendulum bob rises to a height ‘h’ given by,
½ mv^2 = mgh, from which h = v^2/2g = 4/20 =0.2m = 20cm.[Option(b)]
This simple problem changes notably if the collision is inelastic as in the modified question below:
A small metallic dart of mass 20gram moving horizontally with a velocity of 2m/s strikes the wooden bob of a long simple pendulum at rest and gets stuck to it. If the mass of the pendulum bob is 20 gram what is the approximate height up to which the bob is raised?
A steel sphere of mass 20 gram moving horizontally with a velocity of 2m/s collides elastically with the bob of a long simple pendulum at rest. If the mass of the pendulum bob is 20 gram what is the height up to which the bob is raised?
(a) 10cm (b) 20cm (c) 40cm (d) 50cm (e) 60cm
As the collision is elastic and the masses of the sphere and the bob are the same, the sphere will transfer the entire kinetic energy to the bob. (This can be easily proved by equating the total initial momentum and the total initial kinetic energy to the total final momentum and the total final kinetic energy respectively). After receiving the kinetic energy from the sphere, the pendulum bob rises to a height ‘h’ given by,
½ mv^2 = mgh, from which h = v^2/2g = 4/20 =0.2m = 20cm.[Option(b)]
This simple problem changes notably if the collision is inelastic as in the modified question below:
A small metallic dart of mass 20gram moving horizontally with a velocity of 2m/s strikes the wooden bob of a long simple pendulum at rest and gets stuck to it. If the mass of the pendulum bob is 20 gram what is the approximate height up to which the bob is raised?
(a) 2.5cm (b) 5cm (c) 7.5cm (d) 10cm (12.5cm
This is a case of inelastic collision since the dart sticks to the bob and moves with the bob after the impact (Part of the kinetic energy of the dart will be lost in piercing the bob). You should not equate the initial kinetic energy of the dart to the gravitational potential energy of the bob-dart system. The common velocity of the bob-dart system after the impact is to be found from the law of conservation of momentum and the kinetic energy of the bob-dart combination is to be found and equated to the gravitational potential energy.
Equating the initial momentum to the final momentum,we have
0.02×2 + 0=(0.02+0.02)v, from which the common velocity ‘v’of the bob-dart combination works out to be 1m/s. Equating the K.E.and P.E.,
½ (0.02+0.02)×1^2 = (0.02+0.02)×gh.
Taking ‘g’ to be nearly 10, h = 0.05m = 5cm [option(b)]
This is a case of inelastic collision since the dart sticks to the bob and moves with the bob after the impact (Part of the kinetic energy of the dart will be lost in piercing the bob). You should not equate the initial kinetic energy of the dart to the gravitational potential energy of the bob-dart system. The common velocity of the bob-dart system after the impact is to be found from the law of conservation of momentum and the kinetic energy of the bob-dart combination is to be found and equated to the gravitational potential energy.
Equating the initial momentum to the final momentum,we have
0.02×2 + 0=(0.02+0.02)v, from which the common velocity ‘v’of the bob-dart combination works out to be 1m/s. Equating the K.E.and P.E.,
½ (0.02+0.02)×1^2 = (0.02+0.02)×gh.
Taking ‘g’ to be nearly 10, h = 0.05m = 5cm [option(b)]
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