Two bodies of mass m A & m B are moving with velocity V A & V B respectively. We know that V c m = m A + m B m A V A + m B V B . If the two bodies collide :-

KE = KE of COM + KE relative to COM; collision cannot raise KE, and an elastic collision returns the full initial KE - (A) true. KE relative to COM can vanish at best, so minimum KE = 2 1 ( m A + m B ) V c m 2 - (B) true. If V c m = 0 (equal opposite momenta) that minimum is zero - (C) true. Maximum loss = relative-motion KE = 2 1 μ V rel 2 = 2 1 m A + m B m A m B ( V A − V B ) 2 - (D) true. Answer: (A), (B), (C), (D)

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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediummultiple choice

Two bodies of mass $m_{A}$ & $m_{B}$ are moving with velocity $V_{A}$ & $V_{B}$ respectively. We know that $V_{c m} =$ $\frac{m _{A} V _{A} + m _{B} V _{B}}{m _{A} + m _{B}}$ . If the two bodies collide :-

Options:(select one or more)

Answer:

A, B, C, D

Solution:

$KE = KE$ of COM + KE relative to COM; collision cannot raise KE, and an elastic collision returns the full initial KE - (A) true. KE relative to COM can vanish at best, so minimum $KE = \frac{1}{2} (m_{A} + m_{B}) V_{c m}^{2}$ - (B) true. If $V_{c m} = 0$ (equal opposite momenta) that minimum is zero - (C) true. Maximum loss = relative-motion $KE = \frac{1}{2} μ V_{rel}^{2} = \frac{1}{2} \frac{m _{A} m _{B}}{m _{A} + m _{B}} (V_{A} - V_{B})^{2}$ - (D) true. Answer: (A), (B), (C), (D)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 3mℹ️ Source: QB

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