Question

A ball released from a height of 20 m, hits the ground and rebound to a height of 16 m. The coefficient of restitution of collision will be? ​

Answer 1

Answer:

Explanation:

See the attachment

Answer 2

Answer: e=(2/$\sqrt{5}$)≈0.894

Explanation:

e(Coefficient of Restitution is given by:-

e=(Velocity of separation)/(Velocity of approach)

In this case as the wall is stationary velocity of separation becomes the velocity of the ball just after the collision( with which it separates with the wall) and velocity of approach is the velocity of the ball just before it collides the it is the velocity with which it approaches the wall.

So,

e becomes (velocity of the ball after collision/ velocity of the ball before collision)

Find the velocity before and after collision:-

1)Before collision velocity:-

Velocity can be calculated using Kinematic equation:-

Equation:-$v^{2}$=$u^{2}$+2×a×s.

So initial velocity is zero(free fall) and displacement is height=20m

$v^{2}$=0+2×g×20.                      acceleration is g due to freefall

implies v=$\sqrt{40g}$.(Velocity of approach)

2)After collision velocity:-

Using the same Kinematic equation:-

Equation:-$v^{2}$=$u^{2}$+2×a×s.

So final velocity is zero(as it rebounds to that specific height it stops)

and displacement is 16m(height).

0=$u^{2}$+2×(-g)×16.              acceleration is -g due to opposite gravity motion

Taking the negative term to LHS and taking square root.

u=$\sqrt{32g$.(Velocity of separation)

Now the value of (e=u/v)(As per the velocities in formula).

e=$\sqrt{32g}$/$\sqrt{40g}$ so g gets cancelled from both sides and on simplification

e=$\sqrt{32/40}$=$\sqrt{4/5}$(dividing by 8 in numerator and denominator).($\sqrt{4$=2)

Answer:-e=2/$\sqrt{5}$≈0.894.

For more such questions:-https://brainly.in/question/13387617

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