A 0.0351-kg bullet is fired horizontally into a 4.33-kg wooden block attached to
ID: 1978734 • Letter: A
Question
A 0.0351-kg bullet is fired horizontally into a 4.33-kg wooden block attached to one end of a massless, horizontal spring (k = 802 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.159 m. What is the speed of the bullet?Explanation / Answer
let initial vel. of bullet = u
by conservation of momentum,
0.0351*u = (0.0351+4.33)*v, where v =final vel. of bullet-block system at unstrained spring position, just after collision
=> u = 124.3618*v
by conservation of energy,
K.E. +P.E. of bullet-block system at unstrained spring position = K.E. +P.E. of bullet-block system at max. compression of spring, i.e., amplitude of oscillation
=> 0.5*4.3651*(v^2) + 0 = 0 + 0.5*802*(0.159^2) [since P.E. at unstrained position & K.E at amplitude = 0]
=> v^2 = (802/4.3651)*(0.159^2)
=> v = 2.1552
u = 124.3618*2.1552 = 268.02 m/s
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