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Physics · LAWS OF MOTION
हिन्दीAn astronaut of total mass \( 80\text{ kg} \) is at rest outside a spaceship in zero-gravity space. He throws a tool of mass \( 2\text{ kg} \) directly away from himself with a velocity of \( 16\text{ m/s} \) relative to himself. The separation distance between the astronaut and the tool after \( 5\text{ s} \) is:
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Correct answer · Option A
80 m
Explanation
Since no net external force acts on the system, the relative velocity between the astronaut and the tool remains constant at \( v_{\text{rel}} = 16\text{ m/s} \). Distance after \( t = 5\text{ s} \) is \( d = v_{\text{rel}} \times t = 16 \times 5 = 80\text{ m} \).
