Question

As shown in the figure, a 4 kg ball initially moving with horizontal velocity v strikes the wall and rebounds with 75% of its initial speed. It then undergoes head on collision with the 6 kg ball and sticks to it. If both start moving with common speed of 6 m/s, then the value of v is

Answer 1

given initial velocity= v m/sec

after striking wall its velocity= 0.75v

after head on collision with 6kg ball it sticks to it and started moving with 6 m/sec ; law of conservation energy : mv= constant

4x 0.75v + 6v =(4+6)6

9v=60

v=60/9 = 6.6 m/sec

Answer 2

Given, Initial velocity= v m/s

           After hitting the wall, velocity= 75/100 v = 3/4 v m/s

Now, it hits the 6 kg ball which was initially at rest.

Since, total initial momentum = total final momentum

          ∴ m₁u₁+m₂u₂ = m₁v₁+m₂v₂

            4×3/4v+6×0 = 4×6+6×6

( As 4 kg ball hits the 6 kg ball with 75% speed, i.e. 3/4 v)

            3v = 60

             ∴ v = 20 m/s