Answer to Question #161853 in Classical Mechanics for Tony

Question

A 5.00g bullet moving with initial speed of vi = 400 m/s is fired into and passes through a 1.00kg block. The block , initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 900N/m. The block moves 5.00cm to the right after impact before being brought to rest by the spring. Find the speed at which the bullet emerges from the block assuming that energy is conserved.

Answer 1

Let's first find the initial velocity of the block from the law of conservation of energy:

$$KE=PE,$$ $$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2,$$ $$v=\sqrt{\dfrac{kx^2}{m}}=\sqrt{\dfrac{900\ \dfrac{N}{m}\cdot(0.05\ m)^2}{1\ kg}}=1.5\ \dfrac{m}{s}.$$

Then, we can find the speed at which the bullet emerges from the block from the law of conservation of linear momentum:

$$m_{bullet}v_{bullet,i}+m_{block}v_{block,i}=m_{bullet}v_{bullet,f}+m_{block}v_{block,f},$$ $$v_{bullet,f}=\dfrac{m_{bullet}v_{bullet,i}-m_{block}v_{block,f}}{m_{bullet}},$$ $$v_{bullet,f}=\dfrac{5\cdot10^{-3}\ kg\cdot400\ \dfrac{m}{s}-1\ kg\cdot1.5\ \dfrac{m}{s}}{5\cdot10^{-3}\ kg}=100\ \dfrac{m}{s}.$$