A body of mass 2 ,kg collides with a wall wit | vedclass

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(B) The change in momentum $\Delta p$ is given by the final momentum minus the initial momentum.Initial momentum $p_i = m 	imes v = 2 \,kg 	imes 100

Vedclass · Accepted Answer

$2 	imes 10^{4} \,N$

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(B) The change in momentum $\Delta p$ is given by the final momentum minus the initial momentum.Initial momentum $p_i = m 	imes v = 2 \,kg 	imes 100 \,ms^{-1} = 200 \,kg \cdot ms^{-1}$.Since the body rebounds with the same speed in the opposite direction,the final momentum $p_f = m 	imes (-v) = 2 \,kg 	imes (-100 \,ms^{-1}) = -200 \,kg \cdot ms^{-1}$.The change in momentum $\Delta p = p_f - p_i = -200 - 200 = -400 \,kg \cdot ms^{-1}$.The magnitude of the change in momentum is $|\Delta p| = 400 \,kg \cdot ms^{-1}$.The force exerted on the wall is given by Newton's Second Law: $F = \frac{|\Delta p|}{\Delta t}$.Given $\Delta t = 1/50 \,s$,we have $F = \frac{400}{1/50} = 400 	imes 50 = 20,000 \,N = 2 	imes 10^{4} \,N$.

$A$ body of mass $2 \,kg$ collides with a wall with a speed of $100 \,ms^{-1}$ and rebounds with the same speed. If the time of contact is $1/50 \,s$,the force exerted on the wall is

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