Momentum and Impulse III
Applications of Momentum

There are two major applications of Newton's laws of linear momentum.

  1. Recoil of a gun
  2. Jet or rocket explosion

    3. Note: Walking is also an application of linear momentum.

Recoil of a gun

When a bullet is fired from a gun, the gun jerks off or recoils with a velocity, V. This velocity is called the recoil velocity. The recoil of the gun is due to the mass difference between the gun and bullet. The momentum of the system is given by: $$ m_bv_b = m_gV $$ $$ V = \frac{m_bv_b}{m_g} $$ $$ where, $$ $$ m_b = \text{mass of bullet in kg} $$ $$ v_b = \text{velocity of bullet in m/s} $$ $$ m_g = \text{mass of gun in kg } $$ $$ V = \text{Recoil velocity of the gun in m/s} $$ Note: The momentum of the gun is negative due to the recoil of the moves which moves in the opposite direction.


Calculations

Example 1: A gun of mass 3kg fires a bullet of mass 20g with a velocity of 500m/s. Calculate the recoil velocity of the gun (NECO)

Solution

$$ m_g = 3kg $$ $$ m_b = 20g = 0.02kg $$ $$ v_b = 500m/s $$ $$ V = ? $$ $$ V = \frac{m_bv_b}{m_g} $$ $$ V = \frac{0.02 × 500}{3} $$ $$ V = \frac{10}{3} $$ $$ \text{Recoil velocity of the gun} = 3.33m/s $$

Example 2: A machine gun with a mass of 5kg, fires a 50g bullet at a speed of 100m/s. The recoil velocity of the machine gun is (JAMB)

Solution

$$ m_g = 5kg $$ $$ m_b = 50g = 0.05kg $$ $$ v_b = 100m/s $$ $$ V = ? $$ $$ V = \frac{m_bv_b}{m_g} $$ $$ V = \frac{0.05 × 100}{5} $$ $$ \text{Recoil velocity of the gun} = 1m/s $$

Example 3: A gun of mass 2kg fires a bullet of mass 1.6 × 10-2kg due east. If the bullet leaves the nozzle of the gun with a velocity of 150m/s, what is the recoil velocity of the gun? (JAMB)

Solution

$$ m_g = 2kg $$ $$ m_b = 1.6 × 10^{-2} kg $$ $$ v_b = 150m/s $$ $$ V = ? $$ $$ V = \frac{m_bv_b}{m_g} $$ $$ V = \frac{1.6 × 10^{-2} × 150}{2} $$ $$ \text{Recoil velocity} = 1.2m/s \text{ due west} $$

This is because the gun recoils opposite the direction of the bullet (east)

Jet and rocket propulsion

The principle of conservation of linear momentum is used in the propulsion of jet aircraft and rockets. This principle is seen in an inflated balloon. When pierced with a pin in one direction and released, the balloon is seen to move in an opposite direction. The momentum is given by: $$ Ft = mv $$ $$ F = \frac{m}{t}v $$ $$ F = Mv $$ $$ \text{where, }M = \text{mass per second in kg} $$ $$ v = \text{Velocity in m/s} $$ $$ F = \text{Thrust experienced by the rocket} $$


Calculations

Example 1: A rocket burns fuel at a rate of 10kg/s and ejects it a velocity of 5 × 10³ m/s. The thrust experienced by the gas on the rocket is (JAMB)

$$ M = 10kg $$ $$ v = 5 × 10³ m/s $$ $$ F = ? $$ $$ F = Mv $$ $$ F = 10 × 5 × 10³ $$ $$ F = 5 × 10⁴N $$

Example 2: A rocket burns 0.01kg of fuel each second and ejects it as a gas with a velocity of 5000 m/s. What force does the gas exert on the rocket? (JAMB)

Solution

$$ M = 0.01kg $$ $$ v = 5000m/s $$ $$ F = Mv $$ $$ F = 0.01 × 5000 $$ $$ F = 50N $$

Example 3: Fuel was consumed at a steady rate of 5.0 × 10-2 kg per second in a rocket engine in a rocket engine and ejected as a gas with a speed of 4 × 10³ m/s Determine the thrust on the rocket

Solution

$$ M = 5 × 10^{-2}kg $$ $$ v = 4 × 10³m/s $$ $$ F = Mv $$ $$ F = 5 × 10^{-2} × 4 × 10³ $$ $$ F = 200N $$

Summary

Linear momemtum and impulse Calculator







Formulas:
  1. For 1: \( v = \frac{m_bv_b}{m_g} \)

  2. For 2: \( F = mV \)

  3. For 3: \( m = \frac{F}{V} \)
Select the formula to use the calculator. Input the values for the parameters to solve using the calculator. Use the information below as a guide.