Question

A basket ball moving to the right at a speed of 10 m/s makes a head-on collision with a ping-pong ball moving to the left at 15 m/s. What will be the speed of the ping-pong ball after the collision. Ignore the air resistance.

Answer 1

Let us assume mass of basketball = m₁ = 500 g and mass of ping-pong ball = m₂ = 3 g

Let u₁ = 10 m/s be the velocity of basket ball before collision

Let u₂ = -15 m/s be the velocity of ping-pong ball before collision

Let v₁ be the velocity of basket ball after collision and v₂ be the velocity of ping-pong ball

after collision

By conservation of momentum, m₁u₁ + m₂ u₂ = m₁v₁ + m₂v₂

By substituting values in above equation , ( 0.5 10 ) - ( 0.003 15 ) = 0.5 v₁ + 0.003 v₂

Above equation becomes, v₁ + 0.006 v₂ = 9.910 ...................(1)

By conservation of energy ,

.....................(2)

Using eqn.(1) , we rewrite eqn.(2) as

By neglecting (0.006 v₂ )² in comparison with 0.006 v₂² , we get

from above quadratic equation we get , v₂ = 34.85 m/s