Question

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block

Answer 1

Answer:

$K=512J$

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass $m_1$ has an initial velocity $v_{1i}$ and a final velocity $v_{1f}$ and the block of mass $m_2$ has an initial velocity $v_{2i}$ and a final velocity $v_{2f}$ then the initial and final momentum of the system will be:

$p_i=m_1v_{1i}+m_2v_{2i}$

$p_f=m_1v_{1f}+m_2v_{2f}$

Since momentum is conserved, $p_i=p_f$, which means:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We know that the block is brought to rest by the collision, which means $v_{2f}=0m/s$ and leaves us with:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}$

which is the same as:

$v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}$

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

$v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s$

So kinetic energy of the bullet as it emerges from the block will be:

$K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J$