Question

A projectile (mass = 0.15 kg) is fired at and embeds itself in a stationary target (mass = 2.44 kg). With what percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off after being struck?

Answer 1

Answer:

38.6 %

Explanation:

First of all, we have to calculate the final velocity of the block-bullet system. We can apply the law of conservation of momentum:

$m u + M U = (m+M)v$

where

m = 0.15 kg is the mass of the bullet

u is the initial velocity of the bullet

M = 2.44 kg is the mass of the block

U = 0 is the initial velocity of the block (it is at rest)

v is the final velocity of the bullet+block

Solving for v,

$v=\frac{mu}{m+M}$

The total initial kinetic energy of the system is just the kinetic energy of the bullet:

$K = \frac{1}{2}mu^2$

While the final kinetic energy of the block+bullet is:

$K' = \frac{1}{2}(m+M) v^2 = \frac{1}{2}(m+M) \frac{(mu)^2}{(m+M)^2}=\frac{1}{2} \frac{m}{m+M}u^2$

So the fraction of kinetic energy lost is

$\frac{K-K'}{K}=\frac{\frac{1}{2}mu^2 - \frac{1}{2}\frac{m}{m+M}u^2}{\frac{1}{2}mu^2}=\frac{m-\frac{m}{m+M}}{m}=\frac{0.15-\frac{0.15}{0.15+2.44}}{0.15}=0.614$

And so, the fraction of kinetic energy left in the projectile after he flies off the block is

1 - 0.614 = 0.386 = 38.6 %