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Chapter 7: Problem 22

The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is \(0.010 \mathrm{~kg}\) and its velocity is \(+720 \mathrm{~m} / \mathrm{s}\). Her mass (including the gun) is \(51 \mathrm{~kg}\). (a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place? (b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of \(5.0 \times 10^{-4} \mathrm{~kg}\) at a velocity of \(+720 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) -0.141 m/s for the single shot, (b) -0.00706 m/s for the blank cartridge.

Step by step solution

01

Understand the Principle

When a gun is fired, the bullet gains forward momentum, and to conserve momentum, the shooter gains an equal magnitude of backward momentum. This is based on the law of conservation of momentum, which states that the total momentum before and after an event must be equal in a closed system without external forces.
02

Apply Conservation of Momentum (Single Shot)

For the bullet and the woman, the total initial momentum is 0, as both are initially at rest. After firing: \[ m_{bullet} \cdot v_{bullet} + m_{woman} \cdot v_{woman} = 0 \]Where \(m_{bullet} = 0.010 \text{ kg}\), \(v_{bullet} = 720 \text{ m/s}\), \(m_{woman} = 51 \text{ kg}\), and \(v_{woman}\) is the unknown recoil velocity. Solving for \(v_{woman}\):\[ v_{woman} = -\frac{m_{bullet} \cdot v_{bullet}}{m_{woman}} \]
03

Calculating Recoil Velocity (Single Shot)

Substitute the known values:\[ v_{woman} = -\frac{(0.010 \text{ kg}) \cdot (720 \text{ m/s})}{51 \text{ kg}} \]This simplifies to:\[ v_{woman} = -0.141 \text{ m/s} \]
04

Apply Conservation of Momentum (Blank Cartridge)

When firing a blank, the bullet mass changes to \(5.0 \times 10^{-4} \text{ kg}\). Using the formula from Step 2:\[ m'_{bullet} \cdot v_{bullet} + m_{woman} \cdot v_{woman}' = 0 \] Where \(m'_{bullet} = 5.0 \times 10^{-4} \text{ kg}\) and \(v_{woman}'\) is the new recoil velocity.
05

Calculating Recoil Velocity (Blank Cartridge)

Substitute the values into the formula:\[ v_{woman}' = -\frac{(5.0 \times 10^{-4} \text{ kg}) \cdot (720 \text{ m/s})}{51 \text{ kg}} \]This simplifies to:\[ v_{woman}' = -0.00706 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recoil Velocity
When you fire a gun, the bullet shoots forward with high speed. But did you know that this action causes a backwards motion too? That's what we call recoil velocity. It's like the gun pushes you back when you push the bullet forward.

This happens because of the conservation of momentum principle. The total momentum of the bullet and the shooter before firing the gun is zero, since neither is moving initially. After the gun is fired, the bullet and the shooter still need to have the same total momentum, which is zero. This means that the bullet's forward momentum has to be balanced by the shooter鈥檚 backward momentum.

Rekindling a fun fact, this is why characters in movies sometimes dramatically move back when firing a powerful weapon without any support!
Momentum
Momentum can often seem like a tricky concept, but it鈥檚 simply a measure of an object's motion. It depends on both the object鈥檚 mass and velocity. In easier terms, it tells us how hard it is to stop a moving object.

Sometimes, we call momentum the "oomph" that an object has due to its mass and speed. You can think of it as how much "push" it carries. The heavier and faster something is, the more momentum it has.

The bullet and the woman on the oil rig example shows how momentum is transferred. When the bullet gains speed and momentum by being fired, the gun transfers an equal amount of momentum back to the woman. Thus, even though the bullet speeds away, the woman gets moved back too but at a greatly reduced velocity because she's much heavier than the bullet.
Physics Problem Solving
Physics problems often look complicated at first glance, but with a little strategy, they can become much clearer. Let's break it down using our problem as an example.

  • **Understand the problem**: Start by clearly understanding what is given and what you need to find. In our case, we need the recoil velocity of a person firing a gun.
  • **Identify the principles**: The conservation of momentum governs our problem. Recognize how the laws of physics apply here.
  • **Set it up**: Use known formulas (like the momentum conservation formula) and plug in the values you know.
  • **Solve step by step**: Simplify the mathematics. Replace values as shown in the example, and solve for the unknown.

By tackling physics problems methodically鈥攗nderstanding concepts, identifying applicable laws, and solving systematically鈥攖hey become surprisingly manageable. Physics may involve numbers and formulas, but at its heart, it tells a story of how our world works. Focus on the principles, and let the math tell the story!

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Most popular questions from this chapter

ssm Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m}\), a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60\) -kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

At illustrates the physics principles in this problem. An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of \(146 \mathrm{~kg}\). During a space-walk, the unit, which has been completely filled with propellant gas, ejects some gas with a velocity of \(+32 \mathrm{~m} / \mathrm{s}\). As a result, the astronaut recoils with a velocity of \(-0.39 \mathrm{~m} / \mathrm{s}\). After the gas is ejected, the mass of the astronaut (now wearing a partially empty propulsion unit) is \(165 \mathrm{~kg}\). What percentage of the gas propellant in the completely filled propulsion unit was depleted?

At provides a review of the concepts that are important in this problem. For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the platform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that \(0.050 \mathrm{~kg}\) of blood is pushed out of the heart with a velocity of \(+0.25 \mathrm{~m} / \mathrm{s}\) and that the mass of the patient and platform is \(85 \mathrm{~kg}\). Assuming that the patient does not slip with respect to the platform, and that the patient and platform start from rest, determine the recoil velocity.

A wagon is coasting at a speed \(v_{\mathrm{A}}\) along a straight and level road. When ten percent of the wagon's mass is thrown off the wagon, parallel to the ground and in the forward direction, the wagon is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the wagon remains the same, the wagon accelerates to a new speed \(v_{\mathrm{B}}\). Calculate the ratio \(v_{\mathrm{B}} / v_{\mathrm{A}}\).

A volleyball is spiked so that its incoming velocity of \(+4.0 \mathrm{~m} / \mathrm{s}\) is changed to an outgoing velocity of \(-21 \mathrm{~m} / \mathrm{s}\). The mass of the volleyball is \(0.35 \mathrm{~kg}\). What impulse does the player apply to the ball?
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