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Chapter 7: Problem 26

A marksman standing on a motionless railroad car fires a gun into the air at an angle of \(30.0^{\circ}\) from the horizontal. The bullet has a speed of $173 \mathrm{m} / \mathrm{s}\( (relative to the ground) and a mass of \)0.010 \mathrm{kg} .\( The man and car move to the left at a speed of \)1.0 \times 10^{-3} \mathrm{m} / \mathrm{s}$ after he shoots. What is the mass of the man and car? (See the hint in Problem 25.)

Short Answer

Expert verified
Answer: The mass of the man and the railroad car is approximately \(1497.8 \mathrm{kg}\).

Step by step solution

01

Identifying the Given Variables

In this problem the given variables are: - Angle of gunshot: \(30.0^{\circ}\) - Bullet speed relative to the ground: \(173 \mathrm{m} / \mathrm{s}\) - Mass of bullet: \(0.010 \mathrm{kg}\) - Speed of man and car after shooting: \(1.0 \times 10^{-3} \mathrm{m} / \mathrm{s}\) (to the left)
02

Calculate the Initial Momentum of the System

Since the man and the car are initially stationary, the initial momentum of the entire system is 0.
03

Calculate the Horizontal Component of Bullet's Velocity

The bullet is shot at an angle of \(30.0^{\circ}\) from the horizontal. Using trigonometry, we can calculate the horizontal component of the bullet's velocity which is \(v_{bx} = v_{b} \cdot \cos(30^{\circ})\). \(v_{bx} = 173 \text{m/s} \cdot \cos(30^{\circ}) \approx 149.78 \mathrm{m} / \mathrm{s}\)
04

Calculate the Final Momentum of the System

The final momentum of the entire system is the sum of the momenta of the man, car, and bullet after the man shoots. \(P_{total} = P_{bullet} + P_{man + car}\) We know the final momentum of the bullet, \(P_{bullet} = m_{bullet} \cdot v_{bx} = 0.010 \mathrm{kg} \cdot 149.78 \mathrm{m} / \mathrm{s} \approx 1.4978 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) The final momentum of the man and the car after shooting is \(P_{man+car} = (m_{man} + m_{car}) \cdot (-1.0 \times 10^{-3} \mathrm{m} / \mathrm{s})\)
05

Apply the Conservation of Momentum Principle

According to the conservation of momentum principle, the total momentum of the system before and after firing remains constant: \(0 = P_{bullet} + P_{man+car}\)
06

Calculate the Mass of the Man and Car

Solving for \((m_{man} + m_{car})\) in the equation and plugging in the given and calculated values, we get: \((m_{man} + m_{car}) = \frac{- P_{bullet}}{1.0 \times 10^{-3} \mathrm{m} / \mathrm{s}}\) \((m_{man} + m_{car}) \approx \frac{-1.4978 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}}{-1.0 \times 10^{-3} \mathrm{m} / \mathrm{s}} \approx 1497.8 \mathrm{kg}\) So the mass of the man and the car is approximately \(1497.8 \mathrm{kg}\).

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Most popular questions from this chapter

A uranium nucleus (mass 238 u), initially at rest, undergoes radioactive decay. After an alpha particle (mass \(4.0 \mathrm{u}\) ) is emitted, the remaining nucleus is thorium (mass \(234 \mathrm{u}) .\) If the alpha particle is moving at 0.050 times the speed of light, what is the recoil speed of the thorium nucleus? (Note: "u" is a unit of mass; it is not necessary to convert it to kg.)
A car with a mass of \(1700 \mathrm{kg}\) is traveling directly northeast $\left(45^{\circ} \text { between north and east) at a speed of } 14 \mathrm{m} / \mathrm{s}\right.\( \)(31 \mathrm{mph}),$ and collides with a smaller car with a mass of \(1300 \mathrm{kg}\) that is traveling directly south at a speed of \(18 \mathrm{m} / \mathrm{s}\) \((40 \mathrm{mph}) .\) The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?
Block \(A,\) with a mass of \(220 \mathrm{g},\) is traveling north on a frictionless surface with a speed of \(5.0 \mathrm{m} / \mathrm{s} .\) Block \(\mathrm{B}\) with a mass of \(300 \mathrm{g}\) travels west on the same surface until it collides with A. After the collision, the blocks move off together with a velocity of \(3.13 \mathrm{m} / \mathrm{s}\) at an angle of \(42.5^{\circ}\) to the north of west. What was \(\mathrm{B}\) 's speed just before the collision?
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A cannon on a railroad car is facing in a direction parallel to the tracks. It fires a \(98-\mathrm{kg}\) shell at a speed of \(105 \mathrm{m} / \mathrm{s}\) (relative to the ground) at an angle of \(60.0^{\circ}\) above the horizontal. If the cannon plus car have a mass of \(5.0 \times 10^{4} \mathrm{kg},\) what is the recoil speed of the car if it was at rest before the cannon was fired? [Hint: A component of a system's momentum along an axis is conserved if the net external force acting on the system has no component along that axis.]
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