A 5.50 g bullet moving at 664 m/s strikes a 920 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 461 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Question

carson889 · Answer

a)
This is a conservation of momentum question. m_before * v_before = m_after * v_after. 

Let p1 be the momentum of the bullet
Let p2 be the momentum of the block

p1_initial + p2_initial = p1_final + p2_final

$$p1_{i}=m_{i}*v_{i}=\frac{ 5.50g }{ 1000\frac{ g }{ kg } }*664 \frac{ m }{ s } = 3.65 kg\frac{ m }{ s }$$ 
Final momentum of bullet:
$$p1_{f}=m_{f}*v_{f}=\frac{ 5.50g }{ 1000\frac{ g }{ kg } }*461 \frac{ m }{ s } = 2.5355 kg\frac{ m }{ s }$$

We know the initial momentum of the block was zero as its velocity was zero.

Thus we have 3.65 + 0 = 2.5355 + m2*v2
1.1145 = m2*v2
We know m2 = 0.920 kg, so solve for v2
v2 = 1.211 m/s

carson889 · Answer

v_com = (m1v1 + m2v2)  / (m1 + m2) = (3.65 + 0) (kg m/s) / (.920 + 0.0055 ) kg = 3.944 m/s

austinl · Answer

Thank you very much!

carson889 · Answer

You are welcome