Question

a 0.0370 - kg bullet is fired horizontally into a 2.48 - kg wooden block attached to one end of a massless, horizontal spring ($k = 872$ 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 0.244 m. what is the speed of the bullet?

Explanation:

Step1: Apply conservation of mechanical energy

After the collision, the kinetic energy of the bullet - block system is converted into elastic potential energy of the spring.
The kinetic energy of the bullet - block system just after the collision is \(K=\frac{1}{2}(m + M)V^{2}\), and the elastic potential energy of the spring is \(U=\frac{1}{2}kA^{2}\).
By conservation of mechanical energy \(K = U\), so \(\frac{1}{2}(m + M)V^{2}=\frac{1}{2}kA^{2}\).
Solving for \(V\), we get \(V = A\sqrt{\frac{k}{m + M}}\).
Substitute \(m = 0.0370\space kg\), \(M=2.48\space kg\), \(k = 872\space N/m\), \(A = 0.244\space m\) into the formula:

$$
LATEXBLOCK0
$$

Step2: Apply conservation of momentum

Before the collision, the momentum of the system is \(p=mv\) (since the block is at rest, its initial momentum is \(0\)). After the collision, the momentum is \(P=(m + M)V\).
By conservation of momentum \(mv=(m + M)V\).
Solving for \(v\), we get \(v=\frac{(m + M)V}{m}\).
Substitute \(m = 0.0370\space kg\), \(M = 2.48\space kg\), \(V=4.54\space m/s\) into the formula:

$$
LATEXBLOCK1
$$

Answer:

The speed of the bullet is \(309\space m/s\)