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Chapter 7: Problem 11

A gun of mass \(20 \mathrm{~kg}\) has bullet of mass \(0.1 \mathrm{~kg}\) in it. The gun is free to recoil \(804 \mathrm{~J}\) of recoil energy are released on firing the gun. The speed of bullet \(\left(\mathrm{ms}^{-1}\right)\) is (a) \(\sqrt{804 \times 2010}\) (b) \(\sqrt{\frac{2010}{804}}\) (c) \(\sqrt{\frac{804}{2010}}\) (d) \(\sqrt{804 \times 4 \times 10^{3}}\)

Short Answer

Expert verified
The speed of the bullet is option (d): \(\sqrt{804 \times 4 \times 10^{3}}\).

Step by step solution

01

Understand the Problem

We are given a gun and a bullet, and we know the recoil energy of the gun when the bullet is fired. Our task is to find the speed of the bullet using this information.
02

Apply the Conservation of Momentum

Before the bullet is fired, both the gun and the bullet are stationary, thus their combined momentum is zero. After firing, the momentum of the bullet and the recoil of the gun must balance each other: \[ m_g v_g = m_b v_b \]where \( m_g = 20 \text{ kg} \), \( m_b = 0.1 \text{ kg} \), and \( v_g \) and \( v_b \) are the velocities of the gun and bullet respectively after firing.
03

Express Recoil Energy

We know the kinetic energy of recoil (energy of the gun moving backward) is given as 804 J. This kinetic energy can be expressed in terms of the velocity of the gun:\[ \frac{1}{2} m_g v_g^2 = 804 \]
04

Solve for Gun's Velocity

Rearrange the energy equation from Step 3 to find \( v_g \):\[ v_g = \sqrt{\frac{2 \times 804}{m_g}} \]Substituting \( m_g = 20 \text{ kg} \):\[ v_g = \sqrt{\frac{1608}{20}} = \sqrt{80.4} \]
05

Relate Gun's Velocity to Bullet's Velocity

Using the conservation of momentum equation from Step 2:\[ m_b v_b = m_g v_g \]Substituting known values: \[ 0.1 v_b = 20 \times \sqrt{80.4} \]Solving for \( v_b \):\[ v_b = 200 \times \sqrt{80.4} \]
06

Match with Provided Options

Convert \( v_b = 200 \times \sqrt{80.4} \) into a form that matches one of the provided options:\( 80.4 \approx 804 \times 4 \times 10^{-2} \), thus:\[ v_b = \sqrt{804 \times 4 \times 10^{3}} \]Thus, the correct answer is option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Law of Conservation of Momentum
The law of conservation of momentum states that the total momentum of a closed system remains constant, provided no external forces are acting on it. In simpler terms, the momentum before an event is equal to the momentum after the event. This principle is crucial in understanding how bullets and guns work.
Initially, when a gun is not fired yet, both the gun and bullet system has zero momentum because they are not moving. When the bullet is fired, it moves forward, and the gun recoils backward.
To keep momentum conserved, the forward momentum of the bullet is equal to the backward momentum of the gun. This relationship can be expressed with the equation:
  • \( m_g v_g = m_b v_b \)
where \( m_g \) and \( v_g \) are the mass and velocity of the gun, and \( m_b \) and \( v_b \) are the mass and velocity of the bullet.
Recoil Energy
Recoil energy is the energy transferred to a gun when it is fired, causing it to move backwards. This energy is essentially the gun's kinetic energy developed due to recoil.
In our problem, we know that the gun has 804 Joules of recoil energy. The formula for kinetic energy, when applied to recoil energy, is:
  • \( \frac{1}{2} m_g v_g^2 = 804 \)
Recoil energy dictates how much the gun will move or "kick back" when fired. Knowing the recoil energy allows us to calculate the recoil velocity, which is pivotal in understanding the interplay between recoil and forward bullet motion.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. Both the bullet and the gun have kinetic energy when the bullet is fired and the gun recoils.
The general formula for kinetic energy is:
  • \( KE = \frac{1}{2} mv^2 \)
When a bullet is fired, it gains high kinetic energy due to its speed and fired mass. Simultaneously, the gun also gains kinetic energy due to recoil, even if it's not as high as the bullet's energy.
In our scenario, this helps us find the recoil velocity and subsequently the bullet's velocity using our known energy values.
Recoil Velocity
Recoil velocity is the speed at which the gun moves backwards when the bullet is fired. It's a direct result of the conservation of momentum and the given recoil energy.
To find the recoil velocity \( v_g \) of the gun:
  • Rearrange the formula for recoil energy: \( v_g = \sqrt{\frac{2 \times 804}{20}} \)
This calculation gives us the velocity with which the gun pushes back. The recoil velocity, combined with the bullet's mass and speed, ensures the momentum conservation in our system by fulfilling the momentum equation mentioned earlier.

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