Two identical balls A and B are released from the positions shown… - Vidyadip

MCQ

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 $M$ & $N$)

A
$1 : 4$
B
$2 : 1$
✓
$4 : 13$
D
$2 : 5$

✓

Answer

Correct option: C.

$4 : 13$

c
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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