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STD 11 - 5. work,energy,power and collision — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 5. work,energy,power and collision4 Marks
Question
Two identical balls $A$ and $B$ are released from the positions shown in figure. They collide elastically on horizontal portion $MN$. All surfaces are smooth. The ratio of heights attained by $A$ and $B$ after collision will be(Neglect energy loss at $M$ & $N$)
✓

Answer

When the two balls collide with each other as the mass of the two balls is equal, they exchange their velocities on colliding elasticaslly. Let the speed of the ball $\mathrm{B}$ when it reaches back ot the initial position be $v$. then

$4 m g h=\frac{1}{2} m v^{2}+m g h \Rightarrow v=\sqrt{6 g h}$

Height reached by particle $B$ (from highest point the incline)

$H_{B}=\frac{v^{2} \sin ^{2} 60^{\circ}}{2 g}=\frac{9 h}{4},$ total height $=h+\frac{9 h}{4}=\frac{13 h}{4}$

after collision the partcle $A$ reaches the maximum height $=h$

Ratio $=\frac{H_{A}}{H_{B}}=\frac{4}{13}$

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