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Chapter 4: Problem 483

A gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity of $50 \mathrm{~m} / \mathrm{s}$. Because of this the gun is pushed back with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). The mass of the gun is (A) \(1.5 \mathrm{~kg}\) (B) \(3 \mathrm{~kg}\) (C) \(2 \mathrm{~kg}\) (D) \(2.5 \mathrm{~kg}\)

Short Answer

Expert verified
Using the conservation of linear momentum, we set up the equation \(0.04~kg \times 50~m/s + m_{gun} \times (-1~m/s) = 0\). Solving for the mass of the gun, we find that \(m_{gun} = 2~kg\), which corresponds to option (C).

Step by step solution

01

Write the momentum conservation equation

The conservation of linear momentum states that the total momentum before the event (firing the bullet) is equal to the total momentum after the event. Mathematically, this is represented as: \(m_{bullet}v_{bullet} + m_{gun}v_{gun} = (m_{bullet} + m_{gun})v_{final}\)
02

Identify the known quantities

We are given the following information: - Mass of bullet (\(m_{bullet}\)) = \(40~g\) (converted to kg: \(0.04~kg\)) - Velocity of bullet (\(v_{bullet}\)) = \(50~m/s\) - Recoil velocity of the gun (\(v_{gun}\)) = \(-1~m/s\) (Note that we use a negative sign since the gun moves in the opposite direction to the bullet) Our goal is to find the mass of the gun, \(m_{gun}\).
03

Set up an equation for the mass of the gun using momentum conservation

We use the momentum conservation equation from Step 1: \(0.04~kg \times 50~m/s + m_{gun} \times (-1~m/s) = (0.04~kg + m_{gun}) \times 0\)
04

Solve the equation for the mass of the gun

We already know that: \(0.04~kg \times 50~m/s + m_{gun} \times (-1~m/s) = 0\) Now, we can solve for the mass of the gun: \(2~kg \mathrm{~m} / \mathrm{s} - m_{gun}~m/s = 0\) \(m_{gun}~m/s = 2~kg\mathrm{~m} / \mathrm{s}\) \(m_{gun} = 2~kg\)
05

Verify the answer

The answer we found is: \(m_{gun} = 2~kg\) Comparing this to the given options: (A) \(1.5~kg\) (B) \(3~kg\) (C) \(2~kg\) (D) \(2.5~kg\) Our answer matches option (C) which is the correct answer.

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Most popular questions from this chapter

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of \(18 \mathrm{~cm}\) (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(0.4 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \mathrm{~m} / \mathrm{s}\) (D) \(0.6 \mathrm{~m} / \mathrm{s}\)

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