Momentum and CollisionMediumBloom L3
Question
Two identical balls are released from positions as shown. They collide elastically on horizontal surface. Ratio of heights attained by A & B after collision is (All surfaces are smooth, neglect energy loss at M & N)
Options
A.1 : 4
B.2 : 1
C.4 : 13
D.2 :5
Solution
Sol.
u = $\sqrt{2g(4h)}$
Height of A = $\frac{({usin}45{^\circ})^{2}}{2g}$
= $\frac{V^{2} \times \frac{1}{2}}{2g} = \frac{2g(4h)\frac{1}{2}}{2g}$
= 2 h.
v = $\sqrt{2gh}$
max height = $\frac{({Vsin}60{^\circ})^{2}}{2g}$
= $\frac{V^{2} \cdot \frac{3}{4}}{2g}$
= $\frac{(2gh)\frac{3}{4}}{2g}$
= $\frac{3h}{4}$
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