a ball is dropped on a floor of coefficient of restitution 0.5 the percentage loss of energy on rebounding from floor is

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Solution

Dear student,

The kinetic energy before impact is $K . E_{1} = \frac{1}{2} m {v_{1}}^{2}$
The kinetic energy after impact is $K . E_{2} = \frac{1}{2} m {v_{2}}^{2}$
$A s v_{2} = 0.5 v_{1} ∆ E = E_{1} - E_{2} = \frac{1}{2} m {v_{1}}^{2} - \frac{1}{2} m {(\frac{1}{2} v_{1})}^{2} ∆ E = \frac{1}{2} m {v_{1}}^{2} . \frac{3}{4}$

The percentage of loss is 75%.

Regards.

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a ball is dropped on a floor of coefficient of restitution 0.5 the percentage loss of energy on rebounding from floor is

Dear student, The kinetic energy before impact is K.E1=12mv12 The kinetic energy after impact is K.E2=12mv22 As v2=0.5v1∆E=E1-E2=12mv12-12m12v12∆E=12mv12.34 The percentage of loss is 75%. Regards.