If the coefficient of restitution is 0.5, what is the percentage loss of energy on each rebounding of a ball dropped from a height?

Collisions

NEET

25%

50%

67.5%

75%

Solution:

Let $v_1$ be the speed of the ball just before hitting the ground and $v_2$ be the speed just after rebounding. The coefficient of restitution is given by $e = \frac{v_2}{v_1}$ .

Given $e = 0.5$ .

The kinetic energy just before impact is $KE_1 = \frac{1}{2}mv_1^2$ .

The kinetic energy just after impact is $KE_2 = \frac{1}{2}mv_2^2$ .

The loss of energy is $\Delta KE = KE_1 - KE_2 = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2$ .

The percentage loss of energy is $\frac{\Delta KE}{KE_1} \times 100\% = \frac{\frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} \times 100\% = \left(1 - \frac{v_2^2}{v_1^2}\right) \times 100\%$ .

Since $e = \frac{v_2}{v_1}$ , we have $e^2 = \frac{v_2^2}{v_1^2}$ .

Percentage loss of energy $= (1 - e^2) \times 100\%$ .

Substituting $e = 0.5$ , we get:

Percentage loss $= (1 - (0.5)^2) \times 100\% = (1 - 0.25) \times 100\% = 0.75 \times 100\% = 75\%$ .

If the coefficient of restitution is 0.5, what is the percentage loss of energy on each rebounding of a ball dropped from a height?

Solution:

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