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Chapter 7: Problem 60

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the \(2.00-\mathrm{g}\) bullet into a \(2.00-\mathrm{kg}\) wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of \(0.500 \mathrm{~cm}\) above its initial position. What is the velocity of the bullet on leaving the gun's barrel?

Short Answer

Expert verified
Answer: The muzzle velocity of the bullet is approximately 1180 m/s.

Step by step solution

01

List the given information

We're given the following information: - Mass of the bullet (\(m_b\)) = 2.00 g = 0.002 kg (converted to kg) - Mass of the wooden block (\(m_w\)) = 2.00 kg - Initial height of the block (\(h_i\)) = 0 - Maximum height of the block after collision (\(h_f\)) = 0.500 cm = 0.005 m (converted to meters)
02

Conservation of Momentum

Since only the bullet-block system is involved and there is no external force acting on it during the collision, the momentum before the collision is equal to the momentum after the collision. Let's denote the initial velocity of the bullet as \(v_b\) and the initial velocity of both the bullet and block after the collision as \(v_f\). The initial velocity of the wooden block before collision is 0. Using the conservation of momentum, we have: \(m_b v_b + m_w \times 0 = (m_b + m_w) v_f\) Solving for \(v_f\), we get: \(v_f = \frac{m_b}{m_b + m_w} v_b\)
03

Conservation of Mechanical Energy

The mechanical energy is conserved when the block is at its maximum height because there are no external forces acting on the bullet-block system other than gravity. So, the initial kinetic energy of the bullet-block system after the collision is converted to gravitational potential energy when the block reaches its maximum height. We have: \(\frac{1}{2}(m_b + m_w)v_f^2 = (m_b + m_w)gh_f\) Divide both sides by \((m_b + m_w)\): \(\frac{1}{2}v_f^2 = gh_f\) Solving for \(v_f\), we get: \(v_f = \sqrt{2gh_f}\)
04

Finding the initial velocity of the bullet

Now, we can substitute the expression we got for \(v_f\) from Step 2 into the equation from Step 3 and solve for the initial bullet velocity \(v_b\): \(\frac{m_b}{m_b + m_w} v_b = \sqrt{2gh_f}\) Rearrange and solve for \(v_b\): \(v_b = \frac{m_b + m_w}{m_b} \sqrt{2gh_f}\) Finally, plug in the given values: \(v_b = \frac{0.002 + 2.00}{0.002} \sqrt{2 \times 9.81 \times 0.005}\) \(v_b ≈ 1180 \thinspace \text{m/s}\) Thus, the velocity of the bullet on leaving the gun's barrel is approximately \(1180 \thinspace \text{m/s}\).

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