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A ball is dropped on a horizontal surfac...

A ball is dropped on a horizontal surface from height h. If it rebounds upto `h/2` after first collision then coefficient of restitution between ball and surface is

A

`1 sqrt2`

B

`1/2`

C

`1/4`

D

`1/2 + sqrt2`

Text Solution

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The correct Answer is:
To find the coefficient of restitution (e) between a ball and a horizontal surface when the ball is dropped from a height \( h \) and rebounds to a height of \( \frac{h}{2} \), we can follow these steps: ### Step 1: Determine the velocity of the ball just before impact When the ball is dropped from height \( h \), we can use the third equation of motion to find the velocity just before it hits the ground. The equation is: \[ v^2 = u^2 + 2gh \] Here, \( u = 0 \) (initial velocity), \( g \) is the acceleration due to gravity, and \( h \) is the height from which the ball is dropped. Substituting the values, we get: \[ v^2 = 0 + 2gh \implies v = \sqrt{2gh} \] This velocity \( v \) is the velocity of approach (\( V_A \)) just before the ball hits the ground. ### Step 2: Determine the velocity of the ball just after rebound After rebounding to a height of \( \frac{h}{2} \), we can again use the third equation of motion to find the velocity just after the ball leaves the ground. The equation is: \[ v^2 = u^2 - 2gh \] In this case, the final velocity \( v = 0 \) (at the peak of the rebound), and we need to find the initial velocity \( u \) just after the rebound. The height is \( \frac{h}{2} \). Substituting the values, we have: \[ 0 = u^2 - 2g\left(\frac{h}{2}\right) \implies u^2 = gh \implies u = \sqrt{gh} \] This velocity \( u \) is the velocity of separation (\( V_S \)) just after the ball rebounds. ### Step 3: Calculate the coefficient of restitution The coefficient of restitution \( e \) is defined as the ratio of the velocity of separation to the velocity of approach: \[ e = \frac{V_S}{V_A} \] Substituting the values we found: \[ e = \frac{\sqrt{gh}}{\sqrt{2gh}} = \frac{1}{\sqrt{2}} \] ### Final Answer The coefficient of restitution between the ball and the surface is: \[ e = \frac{1}{\sqrt{2}} \] ---

To find the coefficient of restitution (e) between a ball and a horizontal surface when the ball is dropped from a height \( h \) and rebounds to a height of \( \frac{h}{2} \), we can follow these steps: ### Step 1: Determine the velocity of the ball just before impact When the ball is dropped from height \( h \), we can use the third equation of motion to find the velocity just before it hits the ground. The equation is: \[ v^2 = u^2 + 2gh \] ...
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Knowledge Check

  • A ball is dropped on a horizontal surface from height h. If it rebounds upto height (h)/(2) after first collision then coefficient of restitution between ball and surface is

    A
    `(1)/(sqrt2)`
    B
    `(1)/(2)`
    C
    `(1)/(4)`
    D
    `(1)/(2sqrt2)`
  • A ball is dropped from a height h onto a floor and rebounds to a height h//6 . The coefficient of restitution between the ball and the floor is

    A
    `(1)/(2)`
    B
    `(1)/(4)`
    C
    `(2)/(3)`
    D
    `(1)/sqrt(6)`
  • A rubber ball is dropped on the ground from a height of 1m. What is the height to which the ball will rebound, if the coefficient of restitution between the ball and the ground is 0.8?

    A
    0.5m
    B
    0.25n
    C
    0.64m
    D
    0.8m
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