Problem: A bullet of mass 4 grams is shot into a 10 kg mass, which quickly and completely absorbs the bullet. The mass is originally at rest, and is attached at the equilibrium position to a spring whose force constant is 2 Newtons/meter. The mass is observed to move to a maximum displacement of 4 meters from the equilibrium position. Find the velocity of the mass immediately after absorbing the bullet, and the velocity of the bullet immediately before impact. Assume that no dissipative forces, nor any force other than the spring, act on the mass after the collision.
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Solution: At the extreme position, 4 meters from equilibrium, the potential energy of the mass is .5 kx ^ 2 = 16 Joules. Since only the spring forces were acting on the mass between that time and the time the extreme position was attained, no work was done on the system between the instant after the collision and the time at which the maximum displacement was attained. Thus no energy was added to the system during this time.
Therefore the 16 Joules of energy must have been present in the form of kinetic energy just after collision.
It follows that, just after collision, the total mass 10 of the bullet and the larger mass had kinetic energy
initial total KE = 16 Joules.
Solving for the velocity we obtain
v = 1.788 meters/second.
The mass therefore has momentum
momentum = ( 10 kilograms) ( 1.788 meters/second) = 17.88 kilogram meters/second.
Since momentum is conserved in collisions, the total momentum before collision was also 17.88 kilogram meters/second. Before the collision, only the bullet had momentum. If we let v stand for its velocity for collision, we therefore have
( 4 grams)vBullet = ( .004000 kilograms)vBullet = 17.88 kilogram meters/second.
Solving for v, we obtain the bullet's velocity
vBullet = 4470 meters/second.