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Question

Physics Question on work, energy and power

A ball falls from a height of 20 m on the floor and rebounds to a height of 5 m. Time of contact is 0.02 s. Find the acceleration during impact

A

1200 ms−2 1200\, ms^{ - 2}

B

1000 ms−2 1000\, ms^{ - 2}

C

2000 ms−2 2000\, ms^{ - 2}

D

1500 ms−2 1500\, ms^{ - 2}

Answer

1500 ms−2 1500\, ms^{ - 2}

Explanation

Solution

According to Newton's second law of motion, the force acting on a body is equal to the rate of change of momentum during impact F=△p△tF = \frac{ \triangle p }{ \triangle t } Also, F=ma⇒ma=p2−p1△tF = ma \Rightarrow ma = \frac{ p_2 - p_1 }{ \triangle t } ⇒a=(mv2−(−mv1)m△t\Rightarrow a = \frac{ ( mv_2 - ( - mv_1 ) }{ m \triangle t } = v2+v1△t\frac{ v_2 + v_1 }{ \triangle t } So, a=2×10×20+2×10×50.02a = \frac{ \sqrt{ 2 \times 10 \times 20 } + \sqrt { 2\times 10 \times 5 }}{ 0.02 } = 20+100.02=1500 ms−2 \frac{ 20 + 10}{ 0.02 } = 1500 \, ms^{ - 2}