A ball of mass 0.15 ; kg hits the wall with a | vedclass

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(B) The initial momentum of the ball is $\vec{P}_i = m \vec{v}_i = 0.15 	imes 12 \hat{i} = 1.8 \hat{i} \; kg \cdot m/s$.The final momentum of the bal

Vedclass · Accepted Answer

$0.036$

Vedclass · Answer

(B) The initial momentum of the ball is $\vec{P}_i = m \vec{v}_i = 0.15 	imes 12 \hat{i} = 1.8 \hat{i} \; kg \cdot m/s$.The final momentum of the ball after bouncing back is $\vec{P}_f = m \vec{v}_f = 0.15 	imes (-12 \hat{i}) = -1.8 \hat{i} \; kg \cdot m/s$.The change in momentum (impulse) is $\Delta \vec{P} = \vec{P}_f - \vec{P}_i = -1.8 \hat{i} - 1.8 \hat{i} = -3.6 \hat{i} \; kg \cdot m/s$.The magnitude of the change in momentum is $|\Delta \vec{P}| = 3.6 \; kg \cdot m/s$.According to the impulse-momentum theorem,the impulse is equal to the force multiplied by the time duration: $|\Delta \vec{P}| = F \Delta t$.Substituting the given values: $3.6 = 100 	imes \Delta t$.Therefore,$\Delta t = \frac{3.6}{100} = 0.036 \; s$.

$A$ ball of mass $0.15 \; kg$ hits the wall with an initial speed of $12 \; m/s$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 \; N$,calculate the time duration of the contact of the ball with the wall in seconds.

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