Question

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

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

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: \(m_{bullet}v_{bullet} + m_{gun}v_{gun} = (m_{bullet} + m_{gun})v_{final}\)

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, \(m_{gun}\).

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

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

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: \(2~kg \mathrm{~m} / \mathrm{s} - m_{gun}~m/s = 0\) \(m_{gun}~m/s = 2~kg\mathrm{~m} / \mathrm{s}\) \(m_{gun} = 2~kg\)

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.

Key concepts

Understanding Linear Momentum

Linear momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is defined as the product of the object's mass and its velocity. The formula for linear momentum, denoted as \( p \), is given by:

• \( p = mv \)

Here, \( m \) is the mass of the object, and \( v \) is its velocity.
Linear momentum is a vector quantity, which means it has both a magnitude and a direction. This is important to remember, especially when multiple objects are involved, as the direction of motion can significantly affect outcomes.
In problem-solving contexts, understanding linear momentum helps us analyze situations where objects interact, such as collision or recoil, like in the exercise where a gun fires a bullet.

The Concept of Recoil Velocity

Recoil velocity is an application of the conservation of linear momentum. When an object, like a gun, exerts a force to propel another object, such as a bullet, it reacts with a force in the opposite direction, which is felt as recoil.

• This concept is an everyday example observed in firearms, rockets, and even swimming (when pushing off the wall).
• Recoil velocity describes how the gun moves backward as it transfers momentum to the bullet going forward.

In the given exercise, the gun's recoil velocity is \(-1 \mathrm{~m/s}\), indicating it moves in the opposite direction to the bullet with a speed of \(1 \mathrm{~m/s}\).
Remember, the negative sign is crucial since it identifies the opposite direction of movement, aligning with the principle that momentum is conserved across different directions.

Mass and Velocity Relationship

The relationship between mass and velocity is a key theme in understanding momentum. According to the principle of conservation of momentum, the total momentum of a closed system remains constant if no external forces act upon it.
This means that in the gun and bullet system:

• The momentum before firing (when both are at rest) is equal to the momentum after firing.
• Because the bullet and gun move in opposite directions, the mass and velocity of one affect the mass and velocity of the other.

As the bullet with a much smaller mass reaches a high velocity, the heavier gun compensates by moving with a smaller recoil velocity. Thus, even though the bullet travels fast, the gun's recoil is gentle but noticeable due to its larger mass.
The exercise exemplifies this balance, showing how momentum is conserved and distributed between the bullet and the gun.

Problem Solving in Physics

Problem-solving in physics, especially with concepts like momentum, involves understanding the underlying principles and applying them to real-world scenarios. Here’s a simple approach:

• Identify Known and Unknown Variables: Start by listing all given quantities and what you need to find, like in our exercise with the gun and bullet.
• Apply Physics Principles: Use relevant equations, such as momentum conservation (\( m_{bullet}v_{bullet} + m_{gun}v_{gun} = 0 \)), to set up your equations correctly.
• Solve Mathematically: Manipulate the equations to isolate the unknown variable. Double-check units and directions (positive or negative) to ensure they match your overall understanding.

Finally, verify your solution by checking it against given options, as in the exercise, and ensuring consistency with physical laws, demonstrating that the mass of the gun aligns with calculated results.
This structured approach can aid in tackling similar physics problems with confidence.