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Physics · LAWS OF MOTION
हिन्दीA body of mass \( 3\text{ kg} \) changes its velocity vector from \( \vec{v}_1 = (3 \hat{i} + 4 \hat{j})\text{ m/s} \) to \( \vec{v}_2 = (-5 \hat{i} + 10 \hat{j})\text{ m/s} \) during a brief time collision interval \( \Delta t = 0.2\text{ s} \). The magnitude of the average force exerted on the body is:
Correct answer · Option 3
150 N
Explanation
Change in velocity \( \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (-8 \hat{i} + 6 \hat{j})\text{ m/s} \), whose magnitude is \( |\Delta \vec{v}| = 10\text{ m/s} \). Impulse \( |\vec{J}| = m |\Delta \vec{v}| = 3 \times 10 = 30\text{ N s} \). Average force \( F_{\text{avg}} = \frac{30}{0.2} = 150\text{ N} \).
