A rigid ball of mass m strikes a rigid wall a | vedclass

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(C) The impulse imparted by the wall on the ball is equal to the change in momentum of the ball.Let the initial velocity be $\vec{v}_i$ and the final

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$mv$

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(C) The impulse imparted by the wall on the ball is equal to the change in momentum of the ball.Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$.The initial momentum is $\vec{p}_i = m\vec{v}_i$ and the final momentum is $\vec{p}_f = m\vec{v}_f$.Taking the normal to the wall as the $x$-axis,the initial velocity components are $v_{ix} = v \cos 60^\circ$ (towards the wall) and $v_{iy} = v \sin 60^\circ$ (parallel to the wall).The final velocity components are $v_{fx} = -v \cos 60^\circ$ (away from the wall) and $v_{fy} = v \sin 60^\circ$ (parallel to the wall).The change in momentum $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v}_f - \vec{v}_i)$.$\Delta p_x = m(v_{fx} - v_{ix}) = m(-v \cos 60^\circ - v \cos 60^\circ) = -2mv \cos 60^\circ = -2mv(0.5) = -mv$.$\Delta p_y = m(v_{fy} - v_{iy}) = m(v \sin 60^\circ - v \sin 60^\circ) = 0$.The magnitude of the impulse is $|\Delta \vec{p}| = |-mv| = mv$.

$A$ rigid ball of mass $m$ strikes a rigid wall at $60^\circ$ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be

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