Question

Two $500g$blocks of wood are $2.0m$ apart on a frictionless table. A $10g$ bullet is fired at $400m/s$ toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is $6.0m/s$. What is the speed of the second block after the bullet stops in it?

Short answer

The speed of the second block after the bullet stops in it is$1.96m/s$

Step-by-step solution

Step 1. Given information

Mass of block, $m=500g$

Mass of bullet, $m_{bullet}=10g$

The velocity of bullet, $v_{bullet}=400m/s$

Step 2. For the first collision

Apply conservation of principle for collision,

$$\begin{gathered} m_{bullet}\times v_{bullet,intial}+m_{block}\times v_{firstblock,initial}=m_{bullet}\times v_{bullet,final}+m_{block}\times v_{firstblock,final} \\ 0.01kg\times 400m/s+0.5kg\times 0m/s=0.01kg\times v_{bullet,final}+0.5kg\times 6m/s \\ {v}_{bullet,final}=100m/s \end{gathered}$$

Step 3. For second collision

Apply conservation of principle for collision,

$$\begin{gathered} m_{bullet}\times v_{bullet,f}+m_{block}\times v_{secondblock,initial}=(m_{bullet}+m_{block})v_{secondblock,final} \\ 0.01kg\times 100m/s+0.5kg\times 0m/s=(0.01+0.5)v_{secondblock,final} \\ {v}_{secondblock,final}=1.96m/s \end{gathered}$$