Question

A ball is dropped vertically from a height of 3.6 it rebounds from a horizontal surface to a height of 1.6 .find coefficient of restitution..
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Answer 1

Explanation:

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Answer 2

Given:

Height from which the ball is dropped $=3.6m$

Height to which the ball rebounds $=1.6$

To find:

The coefficient of restitution.

Solution:

Step 1

We know, that the ball is released from a height of 3.6 m. Hence, at 3.6 m, the ball has a certain potential energy given by $mgh$ which is converted to the kinetic energy given by $\frac{1}{2}mv^{2}$ during fall.

Initially, the ball is at rest $(u_{1} =0)$ The maximum value of kinetic energy is produced just before the touching of the ball with the ground.

Mathematically,   $mgh=\frac{1}{2}mv^{2}$ , We get the final velocity of the ball as

$v^{2} =2gh$

Substituting the known values in the equation, we get

$v_{1} =\sqrt{2(10)(3.6)}$

$v_{1} =6\sqrt{2}m/s$

Step 2

Now, during the rebound motion of the ball,

It is given that the ball rebounds to a height of $1.6m$ and the velocity of the ball at the highest point that is at 1.6 m will be zero.

$v_{2} ^{2}- u_{2} ^{2} =2as$

We have,  $v_{2} =0m/s$   $;$ $a=-g= -10m/s^{2}$   $;$ $s=1.6m$

$(0)^{2}- u_{2} ^{2} =2(-10)(1.6)$

$u_{2} ^{2} =32$

$u_{2}=4\sqrt{2}m/s$

Step 3

Coefficient of restitution is the ratio of change in velocity of the object after collision with respect to the change in velocity before the collision.

$e=\frac{v_{2}- u_{2} }{v_{1}- u_{1} }$

Substituting the known values, we get

$e=\frac{0-4\sqrt{2} }{6\sqrt{2}-0 }$

$e=-\frac{2}{3}$

We see that the coefficient of restitution in the motion of the ball is negative, hence, the collision is inelastic in nature. Which means that velocity is lost during the collision.

Final answer:

The coefficient of restitution is $-\frac{2}{3}$.