A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest aft…
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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be:
✓ Correct answer: d) 0.25
ExplanationMass of first block (m1) = m Mass of second block (m2) = 4m Initial velocity of first block (u1) = v Since second block (heavier black) is stationary initially, Initial velocity of second block (u2) = 0 Initial momentum of the system After collision, the first block (lighter block) comes to rest, So final velocity of first block (V1) = 0 Final velocity of second block = V2 Final momentum of the system Since there is no external force on the system, so momentum of the system will remain constant Initial momentum (P1) = Final momentum (P2) mv = 4m v2 v2 = v/4 Now use the formula; Thus coefficient of restitution (e) =1 / 4=0.25
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