Ball A moving at 12 m/s collides elastically | vedclass

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(B) In an elastic collision between two objects of equal mass where one is initially at rest,the two objects move at an angle of $90^\circ$ to each ot

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) In an elastic collision between two objects of equal mass where one is initially at rest,the two objects move at an angle of $90^\circ$ to each other after the collision.Let the initial velocity of ball $A$ be $\vec{u}_A = 12 \hat{j} \ m/s$ (assuming the vertical axis is the $y$-axis).Since the collision is elastic and the masses are equal,the conservation of momentum and kinetic energy implies that the velocity vectors after collision $\vec{v}_A$ and $\vec{v}_B$ satisfy $\vec{v}_A \cdot \vec{v}_B = 0$.Given the geometry of the collision,ball $A$ strikes ball $B$ at an angle of $60^\circ$ with the vertical. The component of velocity along the line of impact (normal) is transferred to ball $B$,while the component perpendicular to the line of impact remains with ball $A$.The velocity component of $A$ along the normal (horizontal $x$-axis) is $v_{Ax} = 12 \sin(60^\circ) = 12 	imes \frac{\sqrt{3}}{2} = 6\sqrt{3} \ m/s$.The velocity component of $A$ perpendicular to the normal (vertical $y$-axis) is $v_{Ay} = 12 \cos(60^\circ) = 12 	imes \frac{1}{2} = 6 \ m/s$.In an elastic collision of equal masses,the velocity component along the line of impact is exchanged. Thus,after collision,the $x$-component of $A$'s velocity becomes $0$,and the $y$-component remains $6 \ m/s$.Therefore,the final velocity of ball $A$ is $6 \ m/s$.

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