Collisions and Momentum in Physics

Conservation of Momentum of Systems

When two objects \(A\) and \(B\) collide, the collision can be either (1) elastic or (2) inelastic. Momentum is conserved in all collisions when no external forces are acting. However, kinetic energy is conserved in elastic collisions only.

Inelastic Collisions

In an inelastic collision, deformations of the colliding objects and loss of energy through heat may occur. The heat and deformation energy come from the initial kinetic energy. Momentum is conserved, but kinetic energy is not.

Example 1

On a smooth surface, a soft 100-gram ball \(A\) moving at \(10 \, \text{m/s}\) collides with a stationary 700-gram ball \(B\). After collision, they stick together and move in the same direction as \(A\). What is their final velocity?

Solution

Let \( p_1 \) be the total momentum before collision. \[ p_A = 0.1 \times 10 = 1\,\mathrm{kg\cdot m/s}, \quad p_B = 0.7 \times 0 = 0 \] \[ p_1 = p_A + p_B = 1 \, \mathrm{kg\cdot m/s} \] After collision, combined mass \(m = 0.1 + 0.7 = 0.8 \, \text{kg}\). Let \(v\) be the final velocity. \[ p_2 = 0.8v \] By conservation of momentum: \[ p_1 = p_2 \Rightarrow 1 = 0.8v \] \[ v = 1.25 \, \text{m/s} \]

Elastic Collisions

In elastic collisions, no permanent deformations or heat transfer occur. Both momentum and kinetic energy are conserved.

Example 1

A 100-gram ball \(A\) moving at \(10 \, \text{m/s}\) collides elastically with a 200-gram ball \(B\) moving at \(5 \, \text{m/s}\) in the same direction. After collision, they continue in the same direction with velocities \(v_1\) and \(v_2\). Find \(v_1\) and \(v_2\).

Solution

Total momentum before: \[ p_1 = (0.1 \times 10) + (0.2 \times 5) = 1 + 1 = 2 \, \mathrm{kg\cdot m/s} \] Total momentum after: \[ p_2 = 0.1v_1 + 0.2v_2 \] Conservation of momentum: \[ 2 = 0.1v_1 + 0.2v_2 \quad \text{(1)} \] Kinetic energy before: \[ K_1 = \frac{1}{2}(0.1)(10)^2 + \frac{1}{2}(0.2)(5)^2 = 5 + 2.5 = 7.5 \, \text{J} \] Kinetic energy after: \[ K_2 = \frac{1}{2}(0.1)v_1^2 + \frac{1}{2}(0.2)v_2^2 \] Conservation of kinetic energy: \[ 7.5 = 0.05v_1^2 + 0.1v_2^2 \quad \text{(2)} \] Solve equations (1) and (2). From (1): \[ v_1 = 20 - 2v_2 \] Substitute into (2): \[ 7.5 = 0.05(20 - 2v_2)^2 + 0.1v_2^2 \] Simplify: \[ 3v_2^2 - 40v_2 + 125 = 0 \] Solutions: \[ v_2 = 8.3 \, \text{m/s} \quad \text{or} \quad v_2 = 5 \, \text{m/s} \] Corresponding \(v_1\): \[ v_1 = 20 - 2(8.3) = 3.4 \, \text{m/s} \quad \text{or} \quad v_1 = 20 - 2(5) = 10 \, \text{m/s} \] The second set (\(v_1=10, v_2=5\)) corresponds to the initial condition (no collision). Thus: \[ v_1 = 3.4 \, \text{m/s}, \quad v_2 = 8.3 \, \text{m/s} \]

More References

1 - Higher Level Physics - IB Diploma - Chris Hamper - Pearson
2 - Physics - Raymond A. Serway and Jerry S. Faughn, Holt, Reinehart and Winston