Two balls A and B,of masses M and 2M respecti | vedclass

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(A) Applying the law of conservation of linear momentum before and after the collision:$M(150) + 2M(-v) = M(0) + 2M(v')$ (Taking the direction of ball

Vedclass · Accepted Answer

$37.5$

Vedclass · Answer

(A) Applying the law of conservation of linear momentum before and after the collision:$M(150) + 2M(-v) = M(0) + 2M(v')$ (Taking the direction of ball $A$ as positive)$150 - 2v = 2v'$$v' = 75 - v \dots (1)$Now,the coefficient of restitution $e$ is given by:$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$$1 = \frac{v' - 0}{150 - (-v)}$$150 + v = v' \dots (2)$Equating $(1)$ and $(2)$:$75 - v = 150 + v$$2v = -75$Since we are looking for the speed (magnitude),we consider the relative velocities. Re-evaluating with standard convention: Let $A$ move at $+150$ and $B$ move at $-v$. After collision,$A$ is at $0$ and $B$ moves at $+v'$.$150M - 2Mv = 2Mv' \Rightarrow 75 - v = v'$$e = \frac{v' - 0}{150 - (-v)} = 1 \Rightarrow v' = 150 + v$Solving $75 - v = 150 + v$ gives $v = -37.5$. The magnitude of speed is $37.5 \ m \ s^{-1}$.

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