Question

A 11 gram bullet is fired horizontally into a 100g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having a spring constant of 120 N/m. The bullet becomes embedded in the block. If the bullet block system compresses the spring by a maximum of 55 cm what was the speed of the bullet at impact with the block?

Answer 1

Answer:

The speed of the bullet at impact equals 57.44 m/s

Explanation:

Since the surface is friction less thus no energy shall be lost in the system

Equating the initial kinetic energy of the bullet with the final potential energy of the compressed spring we get

$E_{i}=\frac{1}{2}mv^{2}$

Potential Energy of the compressed spring is given as

$E_{f}=\frac{1}{2}k(\delta x)^{2}$

given

k = 120 N/m

m = 11 grams

x = 55 cm

Equating the 2 energies and solving for velocity we get

$\frac{1}{2}mv^{2}=\frac{1}{2}k(\delta x)^{2}\\\\v^{2}=\frac{k(\delta x)^{2}}{m}\\\\v^{2}=\frac{120\times (0.55)^{2}}{11\times 10^{-3}}\\\\v^{2}=3300\\\\\therefore v=57.44m/s$