Question

Consider a gun of mass $M$ (when unloaded) that fires a shell of mass $m$ with muzzle speed $v$. (That is, the shell's speed relative to the gun is $v$.) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is $v/(1+m/M).$

Short answer

The shell's speed relative to the ground is $\frac{v}{1+\frac{m}{M}}$."

Step-by-step solution

Identify the System and Apply the Law of Conservation of Momentum

In this problem, the system consists of the gun and the shell. Initially, both are at rest, so their total initial momentum is zero. According to the law of conservation of momentum, the momentum before firing is equal to the momentum after firing. We can express this as:$$0=M{v}_g+m{v}_s$$where $v_g$ is the speed of the gun and $v_s$ is the speed of the shell relative to the ground.

Express the Velocity of the Shell

The shell is fired with a muzzle speed of $v$, which is the speed of the shell relative to the gun. This means:$$v_s=v+v_g$$where $v$ is the speed of the shell relative to the gun and $v_g$ is the speed of the gun relative to the ground.

Substitute the Expression into the Momentum Equation

Substitute the expression for $v_s$ from Step 2 into the momentum equation from Step 1:$$0=M{v}_g+m(v+v_g)$$simplifying, we get:$$0=M{v}_g+mv+m{v}_g$$

Solve for the Speed of the Gun

Rearrange the equation from Step 3 to solve for $v_g$:$$M{v}_g+m{v}_g=-mv$$$$(M+m)v_g=-mv$$$$v_g=-\frac{mv}{M+m}$$This is the recoil speed of the gun relative to the ground.

Calculate the Speed of the Shell Relative to the Ground

Substitute the expression for $v_g$ into the expression for $v_s$ to find the speed of the shell relative to the ground:$$v_s=v+v_g$$$$v_s=v-\frac{mv}{M+m}$$This simplifies to:$$v_s=\frac{v(M+m)-mv}{M+m}$$$$v_s=\frac{vM}{M+m}$$$$v_s=\frac{v}{1+\frac{m}{M}}$$Thus, the shell's speed relative to the ground is $\frac{v}{1+\frac{m}{M}}$.

Key concepts

Recoil Velocity

When a gun is fired, the bullet moves forward, and at the same time, the gun moves backward. This backward speed is known as the recoil velocity of the gun. Recoil is a direct consequence of momentum conservation.
In absence of external forces, the total momentum before the gun is fired must equal the total momentum after. Initially, both the gun and the bullet are at rest, so the initial momentum is zero. When the bullet is fired, it moves forward with a certain momentum. To keep the total momentum zero, the gun moves backward with an equal and opposite momentum.

• The momentum of the bullet is given by its mass times its velocity.
• The momentum of the gun is given by its mass times its recoil velocity.

The recoil speed ( v_g ) of the gun is calculated using the equation: $$(M+m)v_g=-mv$$ rearranging, we get: $$v_g=-\frac{mv}{M+m}$$, showing that the gun moves backward with speed inversely proportional to the combined mass.

Muzzle Speed

Muzzle speed is the speed at which a bullet leaves the barrel of a gun. This speed is measured relative to the gun. Muzzle speed is crucial for determining the bullet's impact and range. The higher the muzzle speed, the greater the potential distance the bullet can travel.
The concept of muzzle speed can be better understood through the following points:

• Muzzle speed is the speed of the projectile relative to the gun itself, not the ground.
• It indicates how fast the bullet is traveling the instant it exits the gun barrel.

In this exercise, the muzzle speed ($v$) refers to how fast the shell moves when compared to the gun's own movement, making it a relative measure depending on the point of reference.

Relative Velocity

Understanding relative velocity is key to solving the given problem. When discussing velocities, it's important to specify a reference point. Relative velocity is the velocity of an object as observed from another object's frame of reference, explaining how one object moves compared to another.
In this scenario:

• The muzzle speed is the shell's velocity relative to the gun.
• The velocity of the shell relative to the ground needs both the muzzle speed and gun's recoil velocity.

Combining these velocities: $$v_s=v+v_g$$, where $v_s$ is the velocity of the shell relative to the ground. By substituting the recoil velocity $v_g$, the bullet’s ground speed becomes $\frac{v}{1+\frac{m}{M}}$, using conservation of momentum principles to bridge both perspectives.