Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the velocity of car b after the collision?

Answer 1

m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂;

m₁v₁²/2 + m₂v₂²/2 = m₁u₁²/2 + m₂u²/2

281·2.82 - 209·1.72 = 281u₁ + 209u₂;

281·2.82² + 209·(- 1.72)² = 281u₁² + 209u₂²;

281u₁ + 209u₂ = 432.94 and 281u₁² + 209u₂² = 2852.93

u₁ = 1.05407 - 0.74377u₂;

u₁ = (10.15278 - 0.74377u₂)^1/2

^u₂ = 4.856 m/s

Answer 2

For any collision

Momentum before collision = Momentum after collision

281*2.82 + 209 ( -1.72) = 281V1f + 209V2f

792.42 - 359.48 = 281V1f + 209V2f

432.94 = 281V1f + 209V2f...... Eq1

for Perfectly Elastic collision

velocity of separation /velocity of approach = 1 ( coefficient of restitution being 1)

V2f - V1f = 1.09........ Eq 2

Solving Eq c1 and Eq 2

V1f = 0.4186 m/s

V2f = 1.5086 m/s

So velocity of separation = 1.5086- 0.4186 = 1.09 m/s

Which is also velocity of approach = 2.82 - 1.72 = 1.09 m/s

V2f = Velocity of B after Collision = 1.5086 m/s or + 1.51 m/s