Question

Please solve q 102?

Answer 1

Let,after the collision along the initial pathway of the ball,the wedge will move with velocity `v`,

so,applying law of conservation of momentum,

`m*sqrt(2) = m*0 +2mv` (as after the collision the ball will have no component along its initial diection of motion)

So, `v=1/sqrt(2)`

Now,applying law of conservation of energy,as frictional force is neglected,

`1/2 m (sqrt2)^2 = 1/2 m x^2 +1/2 (2m) (1/sqrt(2))^2` (let,velocity of the ball after the impact is `x`)

So,we get, `2=x^2 +1`

or, `x^2=1`

or, `x=1 ms^-1`