Question

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb was moving at 24.0 m/s to the east. After the explosion, the velocity of the m1 = 115 kg piece is 65.0 m/s to the east. Find the velocity (in m/s) (with a proper sign) of the other piece after the explosion

Answer 1

Answer:

$v_2=-133.17m/s$, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ($p_i$) and after ($p_f$) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$.

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$