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

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 direction is the $y$-axis).Let the final velocities be $\vec{v}_A$ and $\vec{v}_B$.From the conservation of linear momentum: $m\vec{u}_A = m\vec{v}_A + m\vec{v}_B \implies \vec{u}_A = \vec{v}_A + \vec{v}_B$.From the conservation of kinetic energy: $\frac{1}{2}mu_A^2 = \frac{1}{2}mv_A^2 + \frac{1}{2}mv_B^2 \implies u_A^2 = v_A^2 + v_B^2$.Since $\vec{u}_A = \vec{v}_A + \vec{v}_B$, we have $u_A^2 = v_A^2 + v_B^2 + 2\vec{v}_A \cdot \vec{v}_B$. Comparing this with the energy equation, we get $\vec{v}_A \cdot \vec{v}_B = 0$, meaning the balls move at $90^{\circ}$ to each other.From the geometry of the collision, ball $A$ is deflected by $60^{\circ}$ from its original path. Thus, the angle between $\vec{v}_A$ and $\vec{u}_A$ is $60^{\circ}$.Using the projection of velocity: $v_A = u_A \cos(60^{\circ}) = 12 	imes 0.5 = 6 \ m/s$.

Ball $A$ moving at $12 \ m/s$ collides elastically with ball $B$ (initially at rest) as shown in the figure. If both balls have the same mass, what is the final velocity of ball $A$ in $m/s$?

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