Two billiard balls of equal mass 30 , g strik | vedclass

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(B) Impulse $J$ is equal to the change in momentum $\Delta p$. The impulse imparted by the wall is along the normal to the wall (the $X$-direction).Fo

Vedclass · Accepted Answer

$\sqrt{2}: 1$

Vedclass · Answer

(B) Impulse $J$ is equal to the change in momentum $\Delta p$. The impulse imparted by the wall is along the normal to the wall (the $X$-direction).For ball $(a)$, the velocity is perpendicular to the wall. Initial momentum $p_i = mu$ (towards the wall), final momentum $p_f = -mu$ (away from the wall). The change in momentum is $\Delta p_a = |p_f - p_i| = |-mu - mu| = 2mu = J_1$.For ball $(b)$, the velocity makes an angle of $45^{\circ}$ with the normal. The component of velocity perpendicular to the wall is $u \cos 45^{\circ}$. The change in momentum along the normal direction is $\Delta p_b = |(-mu \cos 45^{\circ}) - (mu \cos 45^{\circ})| = 2mu \cos 45^{\circ} = J_2$.The ratio of the magnitudes of impulses is $\frac{J_1}{J_2} = \frac{2mu}{2mu \cos 45^{\circ}} = \frac{1}{\cos 45^{\circ}} = \frac{1}{1/\sqrt{2}} = \sqrt{2} : 1$.

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