A ball is dropped on a smooth horizontal surface from height ‘h’. What will be the height of rebounce after second impact, if coefficient of restitution between ball and surface is ‘e’?
e4h
Understanding Ball Rebound Height with Coefficient of Restitution
This problem involves analyzing the motion of a ball dropped from a height and bouncing off a horizontal surface. The key concept here is the coefficient of restitution, which relates the relative velocities of the objects before and after a collision.
The coefficient of restitution, denoted by 'e', for a collision between a ball and a fixed surface is defined as the ratio of the speed of the ball after the impact to its speed before the impact:
\[e = \frac{\text{Speed after impact}}{\text{Speed before impact}}\]
So, the speed of the ball immediately after impact is $v_{after} = e \times v_{before}$.
First Impact Analysis
The ball is dropped from a height 'h'. Just before the first impact with the surface, its velocity ($v_1$) can be found using the equation of motion $v^2 = u^2 + 2as$. Here, initial velocity $u=0$, acceleration $a=g$, and displacement $s=h$.
\[v_1^2 = 0^2 + 2gh\] \[v_1 = \sqrt{2gh}\]This is the speed of the ball just before the first impact.
Immediately after the first impact, the ball rebounds with a velocity ($v'_1$) given by the coefficient of restitution:
\[v'_1 = e \cdot v_1 = e\sqrt{2gh}\]Now, the ball moves upwards against gravity with this initial velocity $v'_1$. It reaches a maximum height ($h_1$) when its velocity becomes 0. Using $v^2 = u^2 + 2as$ again, with final velocity $v=0$, initial velocity $u=v'_1$, and acceleration $a=-g$ (since it's moving against gravity):
\[0^2 = (v'_1)^2 + 2(-g)h_1\] \[2gh_1 = (v'_1)^2\]Substitute the value of $v'_1$:
\[2gh_1 = (e\sqrt{2gh})^2 = e^2 (2gh)\]So, the height reached after the first rebound is:
\[h_1 = e^2h\]Second Impact Analysis
The ball now falls from the height $h_1 = e^2h$. Just before the second impact with the surface, its velocity ($v_2$) can be found similarly to the first impact. It falls from rest ($u=0$) from height $h_1$ under gravity ($a=g$).
\[v_2^2 = 0^2 + 2gh_1\] \[v_2 = \sqrt{2gh_1} = \sqrt{2g(e^2h)} = e\sqrt{2gh}\]This is the speed of the ball just before the second impact.
Immediately after the second impact, the ball rebounds with a velocity ($v'_2$) given by the coefficient of restitution:
\[v'_2 = e \cdot v_2 = e (e\sqrt{2gh}) = e^2\sqrt{2gh}\]The ball moves upwards with this initial velocity $v'_2$ and reaches a maximum height ($h_2$), which is the height of rebound after the second impact. Using $v^2 = u^2 + 2as$ again, with final velocity $v=0$, initial velocity $u=v'_2$, and acceleration $a=-g$:
\[0^2 = (v'_2)^2 + 2(-g)h_2\] \[2gh_2 = (v'_2)^2\]Substitute the value of $v'_2$:
\[2gh_2 = (e^2\sqrt{2gh})^2 = (e^2)^2 (2gh) = e^4 (2gh)\]So, the height reached after the second rebound is:
\[h_2 = e^4h\]Summary of Rebound Heights
We can see a pattern here:
- Height before 1st impact: \(h = e^0h\)
- Height after 1st impact: \(h_1 = e^2h = e^{2 \times 1}h\)
- Height after 2nd impact: \(h_2 = e^4h = e^{2 \times 2}h\)
In general, the height after the \(n^{th}\) impact will be \(h_n = e^{2n}h\).
Conclusion
Based on our calculation, the height of rebound after the second impact is \(e^4h\).
Let's compare this with the given options:
| Option | Height |
|---|---|
| 1 | \(eh\) |
| 2 | \(e^2h\) |
| 3 | \(e^3h\) |
| 4 | \(e^4h\) |
The calculated height \(e^4h\) matches option 4.
Important Questions from Kinematics and Kinetics
What is the coefficient of restitution (e) for elastic impact?
A body of mass 10 kg moving with a velocity of 1 m/s is acted upon by a force of 50 N for two seconds. The final velocity will be:
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How much force will be exerted by the floor of the lift on a passenger of 80 kg mass when lift is accelerating downward at 0.81 m/s2?
The angular motion of a disc is defined by the relation (θ = 3t + t3), where θ is in radians and t is in seconds. What will be the angular position after 2 seconds?
