A block of mass 0.50 kg moving at 2.00 m/s on a smooth surface strikes another mass of 1.00 kg and they move together. The energy loss during collisio - Physics MCQ AIEEE 2026 | ExamTest.live

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A block of mass 0.50 kg moving at 2.00 m/s on a smooth surface strikes another mass of 1.00 kg and they move together. The energy loss during collision is:

Solution & Step-by-step Explanation

Conservation of momentum: m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V 0.5 \times 2 + 1 \times 0 = (0.5 + 1) V ⟹ 1 = 1.5 V ⟹ V = \frac{2}{3} m/s .Initial Kinetic Energy K_{i} = \frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} (0.5) (2)^{2} = 1.0 J .Final Kinetic Energy K_{f} = \frac{1}{2} (m_{1} + m_{2}) V^{2} = \frac{1}{2} (1.5) (\frac{2}{3})^{2} = \frac{1}{2} (1.5) \frac{4}{9} = \frac{1}{3} J \approx 0.33 J .Energy loss Δ K = K_{i} - K_{f} = 1.0 - 0.33 = 0.67 J .

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A block of mass 0.50 kg moving at 2.00 m/s on a smooth surface strikes another mass of 1.00 kg and they move together. The energy loss during collision is:

0.16 J

1.00 J

0.67 J

0.34 J

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