A marksman standing on a motionless railroad car fires a gun...

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 29.)

Key Concepts

Conservation of Linear Momentum

This principle asserts that in the absence of external forces, the total momentum of a system remains constant. In problems like this one, where the system is initially at rest, the momentum carried by the bullet after being fired must be exactly balanced by the momentum of the marksman and the railroad car in the opposite direction. This allows us to set up an equation equating the bullet's momentum with the recoil momentum, leading to the determination of unknown quantities such as mass.

Vector Decomposition of Momentum

Momentum is a vector quantity, meaning it has both magnitude and direction. When a projectile is fired at an angle, its momentum must be decomposed into horizontal and vertical components. In this scenario, the horizontal component is especially important because it directly pertains to the recoil of the man and car. Decomposing the momentum into components simplifies the problem by allowing focus on the direction in which conservation of momentum is applied.

Reference Frame Consistency

Accurate analysis requires that all measurements be considered within a consistent reference frame. In this problem, the bullet’s speed is given relative to the ground, and the recoil of the man and car is also measured in the same frame. Ensuring that all quantities are expressed relative to the same frame is crucial for correctly applying the conservation laws and obtaining the correct physical relationships.

Transcript

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