Question.
A shell of mass $0.020 \mathrm{~kg}$ is fired by a gun of mass $100 \mathrm{~kg}$. If the muzzle speed of the shell is $80 \mathrm{~m} \mathrm{~s}^{-1}$, what is the recoil speed of the gun?
A shell of mass $0.020 \mathrm{~kg}$ is fired by a gun of mass $100 \mathrm{~kg}$. If the muzzle speed of the shell is $80 \mathrm{~m} \mathrm{~s}^{-1}$, what is the recoil speed of the gun?
solution:
Mass of the gun, M = 100 kg
Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
$\therefore V=\frac{m v}{M}$
$=\frac{0.020 \times 80}{100 \times 1000}=0.016 \mathrm{~m} / \mathrm{s}$
Mass of the gun, M = 100 kg
Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
$\therefore V=\frac{m v}{M}$
$=\frac{0.020 \times 80}{100 \times 1000}=0.016 \mathrm{~m} / \mathrm{s}$