Answer:
![K=512J](https://tex.z-dn.net/?f=K%3D512J)
Explanation:
Since the surface is frictionless, momentum will be conserved. If the bullet of mass
has an initial velocity
and a final velocity
and the block of mass
has an initial velocity
and a final velocity
then the initial and final momentum of the system will be:
![p_i=m_1v_{1i}+m_2v_{2i}](https://tex.z-dn.net/?f=p_i%3Dm_1v_%7B1i%7D%2Bm_2v_%7B2i%7D)
![p_f=m_1v_{1f}+m_2v_{2f}](https://tex.z-dn.net/?f=p_f%3Dm_1v_%7B1f%7D%2Bm_2v_%7B2f%7D)
Since momentum is conserved,
, which means:
![m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%3Dm_1v_%7B1f%7D%2Bm_2v_%7B2f%7D)
We know that the block is brought to rest by the collision, which means
and leaves us with:
![m_1v_{1i}+m_2v_{2i}=m_1v_{1f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%3Dm_1v_%7B1f%7D)
which is the same as:
![v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}](https://tex.z-dn.net/?f=v_%7B1f%7D%3D%5Cfrac%7Bm_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%7D%7Bm_1%7D)
Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:
![v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s](https://tex.z-dn.net/?f=v_%7B1f%7D%3D%5Cfrac%7B%280.01kg%29%282000m%2Fs%29%2B%284kg%29%28-4.2m%2Fs%29%7D%7B0.01kg%7D%3D320m%2Fs)
So kinetic energy of the bullet as it emerges from the block will be:
![K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D%3D%5Cfrac%7B%280.01kg%29%28320m%2Fs%29%5E2%7D%7B2%7D%3D512J)