Question

A 0.010-kg bullet traveling horizontally at \(400.0 \mathrm{m} / \mathrm{s}\) strikes a 4.0 -kg block of wood sitting at the edge of a table. The bullet is lodged into the wood. If the table height is \(1.2 \mathrm{m},\) how far from the table does the block hit the floor?

Short answer

Answer: The block lands approximately 0.49 m away from the edge of the table when it hits the floor.

Step-by-step solution

Calculate the total momentum before and after the collision

According to the conservation of momentum principle, total momentum before the collision is equal to the total momentum after the collision. \(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{f}\) Where \(m_{1} = 0.010\,\text{kg}, v_{1} = 400.0\,\mathrm{m/s}, m_{2} = 4.0\,\text{kg},\) and \(v_{2} = 0.\)

Calculate the final velocity of the block and bullet

Rearrange the equation and solve for \(v_{f}\) \(v_{f} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = \frac{0.010\,\text{kg} \cdot 400.0\,\mathrm{m/s} + 4.0\,\text{kg} \cdot 0}{0.010\,\text{kg} + 4.0\,\text{kg}}\) \(v_{f} \approx 1\,\text{m/s}\)

Determine the time taken for the block to hit the floor

Use the free-fall equation \(h = \frac{1}{2}gt^2\) to find the time taken. \(t = \sqrt{\frac{2h}{g}}\), where \(h = 1.2\,\mathrm{m}\) and \(g = 9.81\,\mathrm{m/s^2}\) \(t \approx 0.49\,\text{s}\)

Calculate the horizontal distance traveled by the block

Use the equation \(d = vt\) to find the distance traveled. \(d = v_{f}t = 1\,\text{m/s} \cdot 0.49\,\text{s}\) \(d \approx 0.49\,\mathrm{m}\) So, the block hits the floor at a distance of approximately \(0.49\,\mathrm{m}\) from the edge of the table.