Center Of Mass Momentum And Collision Question 4
Question 4 - 30 January - Shift 1
As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball $Q$ after collision will be : $(g=10 m / s^{2})$
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Answer: (3)
Solution:
Formula: Conservation of Momentum
The velocities will be interchanged after collision.
Speed of $P$ just before collision $=\sqrt{2 gh}$
$=\sqrt{2 \times 10 \times 0.2}=2 m / s$
BETA
