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Question: A body falling from a height of \(10m\) rebounds from the hard floor. If it loses \(20\% \) energy i...

A body falling from a height of 10m10m rebounds from the hard floor. If it loses 20%20\% energy in the impact, then the coefficient of restitution is .....
(A)0.89(A)0.89
(B)0.56(B)0.56
(C)0.23(C)0.23
(D)0.18(D)0.18

Explanation

Solution

Coefficient of restitution is the ratio of the final velocity to the initial velocity between two objects after the collision takes place. It can also be said that it is a measure of the restitution of a collision between two objects, that is how much energy remains after the two bodies collide with each other.

Complete answer:
According to the question,
The kinetic energy just before the collision will be lost in the form of potential energy.
So, the initial kinetic energy Ki=mgh{K_i} = mgh
On putting the values given to us,
Ki=m×10×10{K_i} = m \times 10 \times 10
Ki=100m........(1){K_i} = 100m........(1)
The final kinetic energy after the impact will be 80%80\% as 20%20\% of the kinetic energy will be lost.
So, the final kinetic energy Kf=Ki×80100{K_f} = {K_i} \times \dfrac{{80}}{{100}}
On putting the value of Ki=100m{K_i} = 100m in the above equation,
Kf=100m×80100{K_f} = 100m \times \dfrac{{80}}{{100}}
Kf=80m......(2){K_f} = 80m......(2)
The expression for kinetic is K=12mv2K = \dfrac{1}{2}m{v^2}
On taking vv on one side and all other terms on the other side,
v2=2Km{v^2} = \dfrac{{2K}}{m}
v=2Km........(3)v = \sqrt {\dfrac{{2K}}{m}} ........(3)
The initial velocity before the impact is vi=2Kim{v_i} = \sqrt {\dfrac{{2{K_i}}}{m}} (from equation 3)
On putting the already known value Ki=100m{K_i} = 100m in the above equation,
vi=2×100mm{v_i} = \sqrt {\dfrac{{2 \times 100m}}{m}}
vi=200{v_i} = \sqrt {200}
vi=102ms......(4){v_i} = 10\sqrt 2 \dfrac{m}{s}......(4)
Similarly, The final velocity before the impact is vf=2Kfm{v_f} = \sqrt {\dfrac{{2{K_f}}}{m}} (from equation 3)
On putting the value of Kf=80m{K_f} = 80m in the above equation,
vf=2×80mm{v_f} = \sqrt {\dfrac{{2 \times 80m}}{m}}
vf=160{v_f} = \sqrt {160}
vf=410ms........(5){v_f} = 4\sqrt {10} \dfrac{m}{s}........(5)
Now,
We know that the coefficient of restitution is e=VfVie = \dfrac{{{V_f}}}{{{V_i}}}
On putting the required values from equation (1) and equation (2), we get,
e=410102e = \dfrac{{4\sqrt {10} }}{{10\sqrt 2 }}
On solving the above equation,
e=0.89e = 0.89
So, the value of the coefficient of restitution comes out to be e=0.89e = 0.89.
So, the correct answer is (A)0.89(A)0.89.

Note: There is a possibility of making mistakes such as understanding the concept of coefficient of restitution as the ratio between the velocities of two bodies before and after the collision of them, which is absolutely wrong. We need to focus only on the relative velocity between the two bodies and not on the velocity of the bodies taking part in the collision. It is important to note that relative velocity is completely different from the velocity of the two bodies.