Three different projectiles,each with the sam | vedclass

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(A) The impulse $\vec{I}$ exerted by the wall on the projectile is equal to the change in momentum of the projectile: $\vec{I} = \Delta \vec{p} = \vec

Vedclass · Accepted Answer

$A > B > C$

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(A) The impulse $\vec{I}$ exerted by the wall on the projectile is equal to the change in momentum of the projectile: $\vec{I} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$.Let the initial direction of the projectile be positive ($+x$ direction).Case $A$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = -mv \hat{i}$. Impulse $I_A = |(-mv) - (mv)| = |-2mv| = 2mv$.Case $B$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = 0$. Impulse $I_B = |0 - mv| = |-mv| = mv$.Case $C$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = \frac{mv}{2} \hat{i}$. Impulse $I_C = |\frac{mv}{2} - mv| = |-\frac{mv}{2}| = \frac{mv}{2}$.Comparing the magnitudes: $2mv > mv > \frac{mv}{2}$,which means $I_A > I_B > I_C$.

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