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Chapter 7: Problem 53

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

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

Step by step solution

01

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 \(v_{2} = 0.\)
02

Calculate the final velocity of the block and bullet

Rearrange the equation and solve for \(v_{f}\) \(v_{f} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = \frac{0.010\,\text{kg} \cdot 400.0\,\mathrm{m/s} + 4.0\,\text{kg} \cdot 0}{0.010\,\text{kg} + 4.0\,\text{kg}}\) \(v_{f} \approx 1\,\text{m/s}\)
03

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 \(g = 9.81\,\mathrm{m/s^2}\) \(t \approx 0.49\,\text{s}\)
04

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.

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Most popular questions from this chapter

A 0.15-kg baseball traveling in a horizontal direction with a speed of $20 \mathrm{m} / \mathrm{s}$ hits a bat and is popped straight up with a speed of \(15 \mathrm{m} / \mathrm{s} .\) (a) What is the change in momentum (magnitude and direction) of the baseball? (b) If the bat was in contact with the ball for 50 ms, what was the average force of the bat on the ball?
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