Consider the situation as shown in the figure | vedclass

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(B) Let the velocity of the wall be $v_w = 20 \ m/s$ (to the right,positive direction).The velocity of the ball before collision is $u = -10 \ m/s$ (t

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(B) Let the velocity of the wall be $v_w = 20 \ m/s$ (to the right,positive direction).The velocity of the ball before collision is $u = -10 \ m/s$ (to the left,negative direction).The mass of the ball is $m = 500 \ g = 0.5 \ kg$.In an elastic collision with a massive wall moving with velocity $v_w$,the velocity of the ball after collision $v$ is given by the formula $v = 2v_w - u$.Substituting the values: $v = 2(20) - (-10) = 40 + 10 = 50 \ m/s$.The change in kinetic energy of the ball is $\Delta KE = KE_f - KE_i = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$.$\Delta KE = \frac{1}{2} 	imes 0.5 	imes (50^2 - (-10)^2) = 0.25 	imes (2500 - 100) = 0.25 	imes 2400 = 600 \ J$.

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