Q. 64 A 100 g ball moving to the right... [FREE SOLUTION]

Chapter 11: Q. 64 (page 290)

A 100 g ball moving to the right at 4.0 m/s catches up and collides with a 400 g ball that is moving to the right at 1.0 m/s. If the collision is perfectly elastic, what are the speed and direction of each ball after the collision?

Short Answer

Expert verified

Velocity of the first ball is 0.8 m/s in backward direction and velocity of the second ball is 2.2 m/s in forward direction.

Step by step solution

Step 1. We need to calculate the speed and direction of each ball after the collision

Mass of the first ball m₁= 100g = 0.1 kg

Initial Velocity of the first ball v_i1 = 4 m/s

Mass of the second ball m₂= 400g = 0.4 kg

Initial Velocity of the second ball v_i2 = 1 m/s

Step 2. Using Law of conservation of momentum and Law of conservation of energy

Collision is elastic therefore momentum and energy is conserved.

Using Law of conservation of Momentum,

$m_{1} v_{i 1} + m_{2} v_{i 2} = m_{1} v_{f 1} + m_{2} v_{f 2} (0.1 k g) (4 m / s) + (0.4 k g) (1 m / s) = (0.1 k g) v_{f 1} + (0.4 k g) v_{f 2} 0.4 k g m / s + 0.4 k g m / s = (0.1 k g) v_{f 1} + (0.4 k g) v_{f 2} 0.8 k g m / s = (0.1 k g) v_{f 1} + (0.4 k g) v_{f 2} v_{f 1} = 8 - 4 v_{f 2}$

Using Law of conservation of Energy,

$\frac{1}{2} m_{1} {v_{i 1}}^{2} + \frac{1}{2} m_{2} {v_{i 2}}^{2} = \frac{1}{2} m_{1} {v_{f 1}}^{2} + \frac{1}{2} m_{2} {v_{f 2}}^{2} m_{1} {v_{i 1}}^{2} + m_{2} {v_{i 2}}^{2} = m_{1} {v_{f 1}}^{2} + m_{2} {v_{f 2}}^{2} (0.1 k g) {(4 m / s)}^{2} + (0.4 k g) {(1 m / s)}^{2} = (0.1 k g) {v_{f 1}}^{2} + (0.4 k g) {v_{f 2}}^{2} (1.6 + 0.4) k g m / s = (0.1 k g) {v_{f 1}}^{2} + (0.4 k g) {v_{f 2}}^{2} 2 k g m / s = (0.1 k g) {(8 - 4 [v_{f 2}])}^{2} + (0.4 k g) {v_{f 2}}^{2} 2 k g m / s = 6.4 + 1.6 {v_{f 2}}^{2} - 6.4 v_{f 2} + 0.4 {v_{f 2}}^{2} 2 k g m / s = 2 {v_{f 2}}^{2} - 6.4 v_{f 2} + 6.4 2 {v_{f 2}}^{2} - 6.4 v_{f 2} + 4.4 = 0 {v_{f 2}}^{2} - 3.2 v_{f 2} + 2.2 = 0 O n s o l v i n g, v_{f 2} = 2.2 m / s o r 1 m / s (1 m / s i s n e g l e c t e d a s v_{i 2} = 1 m / s) v_{f 2} = 2.2 m / s a n d v_{f 1} = 8 - 4 v_{f 2} = 8 - 4 (2.2) m / s = - 0.8 m / s$