Question

From a point on smooth floor of a room, a toy ball is shot to hit a wall. The ball then returns back to the
point of projection. If the time taken by ball in returning is twice the time taken in reaching the wall, find the
coefficient of restitution for collision of ball with wall.​

Answer 1

Answer:

The answer wil be 1/2

Explanation:

According to the diagram let the ball is thrown from point P at angle θ

Let at distance d from P a wall is there

Now let the time taken for touching the wall is t1 and the time taken to come back is t2

There t2 = 2 t1

Now at the time of hitting a force is applied from the wall N

Now before collision the velocity of the ball in x direction vx and in y direction is vy

And after collision the velocity with x axis is v'x and with y axis it will be same.

Now before collision,

as the ball is projected with velocity u

Therefore the components are, ux = u cosθ  and acceleration ax = 0 Now to reach at the point of collision it needs to cover distance d

Therefore displacement , sx = d and time taken t1

Now we can say

vx = ux + axt1

=> vx = ux = u cosθ

we can also say that as acceleration is zero

speed, ux = d/t1 , t1 = d/ucosθ

Now we know coefficient of restitution, e = velocity of separation after collision/velocity of separation before collision

                     e = v'x/vx

v'x = e vx

=> v'x = e x u cosθ

again v'x = d/t2 , t2 = d/v'x = d/ e  u cosθ

Now, t2 = 2 t1

     d/ e  u cosθ = 2 x d/ucosθ

=> e = 1/2

Answer 2

Answer:

e= 1/2

Explanation:

Let d be the distance between point of projection and the wall.

Let the ball hits the wall with velocity u and returns back with velocity v.

Let t and 2t be the time taken hit to and from the wall to the point of projection.

e = velocity of separation/velocity of approach

e = v/u

e = d/2t ×t/d

e = 1/2

Hence Solved!!!!