A ball of mass 0.5 kg moving horizontally at | vedclass

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(A) Linear momentum is a vector quantity.Let the direction towards the wall be positive.Initial momentum $\vec{p}_i = m 	imes 10 \hat{i} = 0.5 	imes

Vedclass · Accepted Answer

$6.0$

Vedclass · Answer

(A) Linear momentum is a vector quantity.Let the direction towards the wall be positive.Initial momentum $\vec{p}_i = m 	imes 10 \hat{i} = 0.5 	imes 10 \hat{i} = 5 \hat{i} \ kg \ m \ s^{-1}$.After rebounding,the ball moves in the opposite direction.Final momentum $\vec{p}_f = -m 	imes v \hat{i} = -0.5 	imes v \hat{i} \ kg \ m \ s^{-1}$.The change in linear momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$.$\Delta \vec{p} = (-0.5v \hat{i}) - (5 \hat{i}) = -(0.5v + 5) \hat{i}$.The magnitude of the change in linear momentum is given as $8.0 \ kg \ m \ s^{-1}$.$|\Delta \vec{p}| = 0.5v + 5 = 8.0$.$0.5v = 8.0 - 5 = 3.0$.$v = \frac{3.0}{0.5} = 6.0 \ m \ s^{-1}$.

$A$ ball of mass $0.5 \ kg$ moving horizontally at $10 \ m \ s^{-1}$ strikes a vertical wall and rebounds with speed $v$. The magnitude of the change in linear momentum is found to be $8.0 \ kg \ m \ s^{-1}$. The magnitude of $v$ is, (in $m \ s^{-1}$)

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