Momentum and Impulse I
Conservation of momentum

The law of conservation of linear momentum states that in a closed or isolated system, the total momentum of the colliding bodies is constant. This can be restated to mean that the total momentum before collision is equal to the total momentum after collision for two colliding bodies acting in a closed system.

$$ m_1u_1+m_2u_2 = m_1v_1+m_2v_2$$

If the bodies move with the same velocity after collision, the common velocity, v is given as: $$ m_1u_1+m_2u_2 = (m_1+m_2)v $$ $$ v = \frac{m_1u_1+m_2u_2}{m_1+m_2} $$ Note: For a stationary body or a body at rest, its initial velocity, u = 0 $$ \therefore v = \frac{m_1u_1}{m_1+m_2} $$

Calculations

Example 1: A body of mass 12kg travelling at 4.2m/s collides with a second body of mass 18kg at rest. Calculate the common velocity if the two bodies coalesce after collision (JAMB)

Solution

$$ m_1 = 12kg $$ $$ u_1 = 4.2m/s $$ $$m_2 = 18kg $$ $$ u_2 = 0 $$ $$ v = ? $$ $$ v = \frac{m_1u_1}{m_1+m_2} $$ $$ v = \frac{12 × 4.2}{12+18} $$ $$ v = \frac{50.4}{30} $$ $$ v = 1.68m/s $$

Example 2: A bullet of mass 0.05kg is fired horizontally into a 10kg block which is free to move. If both bullet and block move with velocity 0.05m/s after the impact, find the velocity with which the bullet hit the block.

Solution

$$ m_1 = 0.05kg $$ $$ m_2 = 10kg $$ $$ u_1 = ? $$ $$ u_2 = 0 $$ $$ v = 0.5m/s $$ $$ v = \frac{m_1u_1}{m_1+m_2} $$ $$ 0.5 = \frac{0.05u_1}{10.05} $$ $$ u_1 = \frac{10.05 × 0.5}{0.05} $$ $$ u_1 = 100.5m/s $$

Example 3: A body (P) of mass 5kg moving with a velocity of 30m/s collides with another body, Q moving in opposite direction with a velocity of 20m/s. If both bodies now move in the direction of p at a velocity of 10m/s, calculate the mass of Q.

Solution

$$ m_1 = 5kg $$ $$ u_1 = 30m/s $$ $$ m_2 = ? $$ $$ u_2 = -20m/s \text{ (opposite direction)} $$ $$ v = 10m/s $$ $$ m_1u_1 + m_2u_2 = (m_1+m_2)v $$ $$ 5 × 30 + (-20m_2) = (5 + m_2)10 $$ $$ 150 - 20m_2 = 50 + 10m_2 $$ $$ 150 - 50 = 10m_2 + 20m_2 $$ $$ 100 = 30m_2 $$ $$ m_2 = \frac{100}{30} $$ $$ \text{mass of Q} = 3.33kg $$

Example 4: A ball of mass 100g travelling with a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in the same direction. If they stick together, what is their common velocity? (NECO)

Solution

$$ m_1 = 100g = 0.1kg $$ $$ m_2 = 400g = 0.4kg $$ $$ u_1 = 100m/s $$ $$ u_2 = 50m/s $$ $$ v = ? $$ $$ v = \frac{m_1u_1+m_2u_2}{m_1+m_2} $$ $$ v = \frac{0.1×100 + 0.4 × 50}{0.1+0.4} $$ $$ v = \frac{30}{0.5} $$ $$ \text{common velocity} = 60m/s $$

Example 5: A ball P of mass 0.25kg, loses one third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2m/s in the original direction of Calculate the initial velocity of P. (WAEC)

Solution

$$ m_1 = 0.25kg $$ $$ m_2 = 0.25kg \text{ (identical ball)} $$ $$ u_1 = u $$ $$ u_2 = 0 $$ $$ v_1 = u - \frac{1}{3} $$ $$ v_1 = \frac{2u}{3} \text{ m/s}$$ $$ v_2 = 2m/s $$

At constant mass; $$ m(u_1+u_2) = m(v_1+v_2) $$ $$ u_1 + u_2 = v_1 + v_2 $$ $$ u + 0 = \frac{2u}{3}+ 2 $$ $$ u = \frac{2u + 6}{3} $$ $$\text{cross multiply} $$ $$ 3u = 2u + 6 $$ $$ u = 6m/s $$ $$ \text{Initial velocity of p} = 6m/s $$

Common Velocity Calculator







Formula: $$ v = \frac{m_1u_1+m_2u_2}{m_1+m_2} $$