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

38.6 %

Step-by-step 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=(mu)/(m+M)`

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

`K = (1)/(2)mu^2`

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

`K' = (1)/(2)(m+M) v^2 = (1)/(2)(m+M) ((mu)^2)/((m+M)^2)=(1)/(2) (m)/(m+M)u^2`

So the fraction of kinetic energy lost is

`(K-K')/(K)=((1)/(2)mu^2 - (1)/(2)(m)/(m+M)u^2)/((1)/(2)mu^2)=(m-(m)/(m+M))/(m)=(0.15-(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 %