The collision of two bodies can either be elastic or inelastic.

Elastic collision: The collision of two bodies is said to be perfectly elastic if the kinetic energy before collision is equal to the kinetic energy after collision.An example of a nearly perfectly elastic collision is the collision of gases.

K.E before collision = K.E after collision

Inelastic collisions: The collision of two bodies is said to be perfectly inelastic if the kinetic energy before collision is greater than the kinetic energy after collision. in this type of collision, the colliding bodies move with a common velocity after collision.

Example 1: A lead bullet of mass 0.05kg is fired with a velocity of 200m/s into a lead block of mass 0.95kg. Given that the lead block can move freely, the final K.E after impact is? (JAMB)

The collision here is inelastic $$ m_1 = 0.05kg $$ $$ m_2 = 0.95kg $$ $$ u_1 = 200m/s $$ $$ u_2 = 0 $$ $$ v = ? $$ $$ \text{K.E after collision} = \frac{1}{2}(m_1+m_2)v² $$ To get the value for common velocity, v $$ v = \frac{m_1u_1}{m_1+m_2} $$ $$ v = \frac{0.05 × 200}{0.05 + 0.95} $$ $$ v = 10m/s $$ $$ \text{KE after collision} = \frac{1}{2}(m_1+m_2)v² $$ $$ = \frac{1}{2}(0.05+0.95) × 10² $$ $$ = \frac{1}{2} × 1 × 100 $$ $$ \text{KE after collision} = 50J $$

Example 2: Two bodies A and B of masses 4kg and 2kg move towards each other with velocities 3m/s and 2m/s and collide. If the collision is perfectly inelastic, find the velocity of the two bodies after collision. Find the total kinetic energy before and after collision; hence calculate the loss in kinetic energy.

1. Since the collision is perfectly inelastic, both bodies move with a common velocity, v $$ v = \frac{m_1u_1+m_2u_2}{m_1 + m_2} $$ $$ v = \frac{4×3+(2×(-2))}{4+2} $$ $$ v = \frac{12 -4}{6} $$ $$ v = \frac{8}{6} = 1.33m/s $$
2. The kinetic energies is given by: $$ K.E_b = \frac{1}{2}m_1u_1² + \frac{1}{2}m_2u_2² $$ $$ = \frac{1}{2} × 4 × 3² + \frac{1}{2} × 2 × 2² $$ $$ = 18 + 4 $$ $$ \text{KE before collision} = 22J $$
   $$ KE_a = \frac{1}{2}(m_1+m_2)v² $$ $$ = \frac{1}{2}(4+2)× 1.33² $$ $$ = 3 × 1.33² $$ $$ \text{KE after collision} = 5.31J $$
3. The loss in kinetic energy is given by: $$\text{loss in K.E} = K.E_b - K.E_a $$ $$ = 22 - 5.31 $$ $$ \text{Loss in K.E} = 16.69J $$

Example 3: A tractor of mass 5.0 × 10³kg is used to tow a car of mass 2.5 × 10³kg. The tractor moved with a speed of 3m/s just before the towing rope went taut. Calculate the:

1. Speed of the tractor immediately the rope becomes taut
2. loss in K.E of the system just after the car has started moving
3. Impulse in the rope when it jerks the car into motion
(WAEC)

1. The common velocity can be determined by; $$ v = \frac{m_1u_1}{m_1+m_2} $$ $$ v = \frac{5×10³×3}{5×10³+2.5×10³} $$ $$ v = \frac{15000}{7500} = 2m/s $$
2. Loss in kinetic energy is determined by: $$ \text{loss in K.E} = K.E_b - K.E_a $$ $$ K.E_b = \frac{1}{2}m_1u_1² + \frac{1}{2}m_2u_2² $$ $$ \text{Since v = 0} $$ $$ K.E_b = \frac{1}{2} × 5000 × 3² $$ $$ K.E_b = 22500J $$
   $$ K.E_a = \frac{1}{2}{(m_1+m_2)}v² $$ $$ K.E_a = \frac{1}{2}{(5000+2500)}× 2² $$ $$ K.E_a = 7500 × 2 $$ $$ K.E_a = 15000J $$
   $$ \text{loss in K.E} = 22500 - 15000 $$ $$ \text{loss in K.E} = 7500J $$
3. The impulse is given by: $$ impulse = \text{change in momentum} $$ $$ ∆p = p_a - p_b $$ $$ where, $$ $$ p_a = \text{momentum after collision} $$ $$ p_b = \text{momentum before collision} $$ $$ p_a = (m_1+m_2)v $$ $$ p_a = (5000+2500) × 2 $$ $$ p_a = 15000kgm/s $$
   $$ p_b = m_1u_1+ m_2u_2 $$ $$ p_b = 5000 × 3 + 0 $$ $$ p_b = 15000kgm/s $$
   $$ ∆p = 15000 - 15000 $$ $$ ∆p = impulse = 0Ns $$