Answer:
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](https://img.qammunity.org/2020/formulas/physics/high-school/pofq0z4owxinpbhgt8hb9w72tgtzii9s3o.png)
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)](https://img.qammunity.org/2020/formulas/physics/high-school/vz5ecry6256o2ckgd5268s6grub0jltyu3.png)
The total initial kinetic energy of the system is just the kinetic energy of the bullet:
![K = (1)/(2)mu^2](https://img.qammunity.org/2020/formulas/physics/high-school/kg0j37di91irm0x8rueb58o5kxbw4y895t.png)
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](https://img.qammunity.org/2020/formulas/physics/high-school/99m50m0pa6kosadisof3uwnl9titp7lfh0.png)
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](https://img.qammunity.org/2020/formulas/physics/high-school/37v7aysokaei1136qhv4t414a51z3io8zj.png)
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 %