Two blocks of mass, 1 kg and 2 kg are moving with velocities of \hat{i} and 4 \hat{i} respectively. Statement - 1 : The kinetic energy of system appears minimum relative to an observer moving with a velocity of 3 \hat{i} m/s. Statement - 2 : The kinetic energy of any 2 block system is \frac{1}{2} MV_{c}^{2} + \frac{1}{2} μV_{rel}^{2} where symbols have their usual meanings.

Engineering

Physics

Center of Mass

Collision

Momentum and Energy Conservation for System of Particles

Question

Two blocks of mass, 1 kg and 2 kg are moving with velocities of $\hat{i}$ and $4 \hat{i}$ respectively.
Statement - 1 : The kinetic energy of system appears minimum relative to an observer moving with a velocity of $3 \hat{i}$ m/s.
Statement - 2 : The kinetic energy of any 2 block system is $\frac{1}{2} {MV}_{c}^{2} + \frac{1}{2} {μV}_{rel}^{2}$ where symbols have their usual meanings.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is false.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

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Statement-1 claims the system's kinetic energy is minimized for an observer moving at 3 m/s. The center of mass velocity is calculated as:

$V_{c} = \frac{(1 \times 1) + (2 \times 4)}{1 + 2} = 3 \hat{i}$ m/s.

Statement-2 is the standard formula for total kinetic energy: $K = \frac{1}{2} M V_{c}^{2} + \frac{1}{2} μ v_{r e l}^{2}$ . The internal term is frame-independent. The center of mass kinetic energy is minimized (zero) when the observer moves with $V_{c}$ , making the total kinetic energy minimum. Therefore, both statements are true, and statement-2 correctly explains statement-1.

Final answer: Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.