Find the velocity of a mass of 4.5 kg, initially at rest, immediately after absorbing a bullet of mass 32 grams moving at an unknown initial velocity if the mass is attached at its equilibrium position to a spring with restoring force constant 692 N / m, if the system is observed to displace to .045 meters amplitude with respect to the equilibrium position. Determine also the velocity of the bullet immediately before impact.
Assume that no dissipative forces act on the mass after the collision.
At the extreme position, 32 meters from equilibrium, the potential energy of the mass is .5 kx ^ 2 = 7.999 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 7.999 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 4.532 of the bullet and the larger mass had kinetic energy
initial total KE = 7.999 Joules.
Solving for the velocity we obtain
v = 3.53 meters/second.
The mass therefore has momentum
momentum = ( 4.532 kilograms) ( 3.53 meters/second) = 15.99 kilogram meters/second.
Since momentum is conserved in collisions, the total momentum before collision was also 15.99 kilogram meters/second. Before the collision, only the bullet had momentum. If we let v stand for its velocity for collision, we therefore have
( 32 grams)* vBullet = ( .032 kilograms) * vBullet = 15.99 kilogram meters/second.
Solving for vBullet, we obtain the bullet's velocity
"vBullet = 500 meters/second.