Question

A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through? What is the kinetic energy loss of the system?

Answer 1

Answer:

speed of block is 0.665 m/s

Kinetic energy loss is 237.16 J

Explanation:

Here we can use momentum conservation as there is no external force on the system in horizontal direction

so here we will have

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

$0.022(210) + 2(0) = 0.022(150) + 2 v$

$4.62 = 3.3 = 2 v$

$v = \frac{4.63 - 3.3}{2}$

$v = 0.665 m/s$

Now kinetic energy loss in the system is given as

$Loss = KE_i - KE_f$

$Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)$

$Loss = 237.16 J$