Question

In a ballistics test, a $25g$ bullet traveling horizontally at $1200m/s$ goes through a 30-cm-thick 350 kg stationary target and emerges with a speed of $900m/s$. The target is free to slide on a smooth horizontal surface. What is the target’s speed just after the bullet emerges?

Short answer

The target's s speed just after the bullet emerges is$0.02m/s$

Step-by-step solution

Step 1. Given information

Mass of bullet, $m_{bullet}=25g$

Initial velocity of the bullet, role="math" localid="1648141886819" $v_{bullet,i}=1200m/s$

Mass of stationary target, $m_{target}=350kg$

Final Velocity of the bullet,$v_{bullet,f}=900m/s$

Step 2. Analyze and apply

Applying the conservation of momentum,

$\begin{array}{l}{m}_{\text{bullet}}{v}_{\text{bullet,initial}}+m_{\text{target}}{v}_{\text{target,initial}}=m_{\text{bullet}}{v}_{\text{bullet,final}}+m_{\text{target}}{v}_{\text{target,inal}}\\ 0.025\times kg1200m/s+350kg\times 0m/s=0.025kg\times 900m/s+350kg\times v_{\text{target,final}}\\ \Rightarrow v_{\text{target,final}}=0.02m/s\end{array}$