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:
[tex]m u + M U = (m+M)v[/tex]
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,
[tex]v=\frac{mu}{m+M}[/tex]
The total initial kinetic energy of the system is just the kinetic energy of the bullet:
[tex]K = \frac{1}{2}mu^2[/tex]
While the final kinetic energy of the block+bullet is:
[tex]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[/tex]
So the fraction of kinetic energy lost is
[tex]\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[/tex]
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 %
