A ball of mass 0.5, kg moving with a velocity | vedclass

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Question

(A) The average force $F$ exerted on the ball is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.Initial momentum $p_i = m

Vedclass · Accepted Answer

$2000$

Vedclass · Answer

(A) The average force $F$ exerted on the ball is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.Initial momentum $p_i = m 	imes v = 0.5 	imes 2 = 1\, kg \cdot m/s$.Final momentum $p_f = m 	imes (-v) = 0.5 	imes (-2) = -1\, kg \cdot m/s$ (since it bounces back).Change in momentum $\Delta p = p_f - p_i = -1 - 1 = -2\, kg \cdot m/s$.The magnitude of the change in momentum is $|\Delta p| = 2\, kg \cdot m/s$.Given time of contact $\Delta t = 1\, ms = 10^{-3}\, s$.Average force $F = \frac{|\Delta p|}{\Delta t} = \frac{2}{10^{-3}} = 2000\, N$.

$A$ ball of mass $0.5\, kg$ moving with a velocity of $2\, m/s$ strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is $1\, ms$,the average force exerted by the wall on the ball is ........ $N$.

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