Question

a 0.0456 - kg bullet is fired horizontally into a 2.00 - kg wooden block attached to one end of a massless, horizontal spring (k = 843 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 0.163 m. what is the speed of the bullet?

Explanation:

Step1: Use conservation of mechanical energy

After the collision, the kinetic energy of the bullet - block system is converted into elastic potential energy of the spring.
By the conservation of mechanical energy \(E_{k}=\frac{1}{2}(m + M)V^{2}=\frac{1}{2}kx^{2}\), where \(m = 0.0456\space kg\) (mass of bullet), \(M=2.00\space kg\) (mass of block), \(k = 843\space N/m\) (spring constant), \(x = 0.163\space m\) (amplitude).
We can solve for the velocity \(V\) of the bullet - block system immediately after the collision:

$$
LATEXBLOCK0
$$

Substitute the values: \(V=\sqrt{\frac{843\times(0.163)^{2}}{0.0456+ 2.00}}\)

$$
LATEXBLOCK1
$$

Step2: Use conservation of momentum

Before the collision, the momentum of the system is \(p_{i}=mv\) (only the bullet is moving). After the collision, the momentum is \(p_{f}=(m + M)V\) (completely inelastic collision).
By the law of conservation of momentum \(p_{i}=p_{f}\), so \(mv=(m + M)V\)
We can solve for the initial velocity \(v\) of the bullet: \(v=\frac{(m + M)V}{m}\)
Substitute \(m = 0.0456\space kg\), \(M = 2.00\space kg\) and \(V\approx3.31\space m/s\)

$$
LATEXBLOCK2
$$

Answer:

The speed of the bullet is approximately \(149\space m/s\)