A point mass of 1 , kg collides elastically w | vedclass

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(A) Let $u$ be the initial velocity of the $1 \, kg$ mass and $v$ be the final velocity of the $5 \, kg$ mass.By conservation of linear momentum: $1 \

Vedclass · Accepted Answer

$(A, C)$

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(A) Let $u$ be the initial velocity of the $1 \, kg$ mass and $v$ be the final velocity of the $5 \, kg$ mass.By conservation of linear momentum: $1 \cdot u + 5 \cdot 0 = 1 \cdot (-2) + 5 \cdot v \implies u = 5v - 2$ ... $(i)$By Newton's experimental law of collision (for elastic collision, coefficient of restitution $e = 1$): $v - (-2) = 1 \cdot (u - 0) \implies u = v + 2$ ... $(ii)$Solving $(i)$ and $(ii)$: $5v - 2 = v + 2 \implies 4v = 4 \implies v = 1 \, m/s$. Then $u = 3 \, m/s$.$(A)$ Total momentum $P = 1 \cdot u = 1 \cdot 3 = 3 \, kg \cdot m/s$. (Correct)$(B)$ Momentum of $5 \, kg$ mass $= 5 \cdot v = 5 \cdot 1 = 5 \, kg \cdot m/s$. (Incorrect)$(C)$ Velocity of centre of mass $v_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{1 \cdot 3 + 5 \cdot 0}{1 + 5} = 0.5 \, m/s$. Kinetic energy of $CM$ $= \frac{1}{2} (m_1 + m_2) v_{cm}^2 = \frac{1}{2} \cdot 6 \cdot (0.5)^2 = 3 \cdot 0.25 = 0.75 \, J$. (Correct)$(D)$ Total kinetic energy $= \frac{1}{2} \cdot 1 \cdot (-2)^2 + \frac{1}{2} \cdot 5 \cdot (1)^2 = 2 + 2.5 = 4.5 \, J$. (Incorrect)Thus, statements $(A)$ and $(C)$ are correct.

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