A 3.00-g bullet traveling horizontally at 400 m/s hits a 3.00-kg wooden block, which is initially at rest on a smooth horizontal table.The bullet buries itself in the block without passing through.The speed of the block after the collision is:
A)1.33 m/s
B)0.40 m/s
C)12.0 m/s
D)12.6 m/s
E)40.0 m/s

Question

Quizplus · Accepted Answer

The conservation of momentum principle applies here. The total momentum before the collision equals the total momentum after the collision. The bullet's mass is 0.003 kg, and its initial velocity is 400 m/s, so its initial momentum is $0.003 \, \text{kg} \times 400 \, \text{m/s} = 1.2 \, \text{kg} \cdot \text{m/s}$. The block's mass is 3.00 kg, and it is initially at rest, so its initial momentum is 0. After the collision, the bullet and block move together with a combined mass of $3.003 \, \text{kg}$. Let $v$ be their final velocity. The conservation of momentum gives $1.2 = (3.003) \cdot v$, solving for $v$ gives $v \approx 0.40 \, \text{m/s}$.

A 300-G Bullet Traveling Horizontally at 400 M/s Hits a 3