Question

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

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

Step-by-step solution

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)

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.

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 \(v_{bx} = v_{b} \cdot \cos(30^{\circ})\). \(v_{bx} = 173 \text{m/s} \cdot \cos(30^{\circ}) \approx 149.78 \mathrm{m} / \mathrm{s}\)

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 \(P_{man+car} = (m_{man} + m_{car}) \cdot (-1.0 \times 10^{-3} \mathrm{m} / \mathrm{s})\)

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: \(0 = P_{bullet} + P_{man+car}\)

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 \(1497.8 \mathrm{kg}\).