Two identical balls A and B are released from the positions shown in the figure. They collide elastically on horizontal portion MN. The ratio of the respective heights attained by A and B after collision will be (neglect friction)

Question

Two identical balls A and B are released from the positions shown in the figure. They collide elastically on horizontal portion MN. The ratio of the respective heights attained by A and B after collision will be (neglect friction) (A) 1: 4 (B) 2: 5 (C) 2: 1 (D) 4: 13

Tardigrade Admin · Accepted Answer

Since collision is elastic hence velocities will be exchanged. After collision vA=√2gh vB=√8gh HA=(v2A/2g)=h HB = (v2B/2g) = 4h (HA/HB)=(1/4)

Q. Two identical balls A and B are released from the positions shown in the figure. They collide elastically on horizontal portion MN. The ratio of the respective heights attained by A and B after collision will be (neglect friction)

1 : 4

2 : 5

2 : 1

4 : 13

Since collision is elastic hence velocities will be exchanged.
After collision $v_{A} = 2 g h$
$v_{B} = 8 g h$
$H_{A} = \frac{v _{A}^{2}}{2 g} = h$
$H_{B} = \frac{v _{B}^{2}}{2 g} = 4 h$
$\frac{H _{A}}{H _{B}} = \frac{1}{4}$

Q. Two identical balls A and B are released from the positions shown in the figure. They collide elastically on horizontal portion MN. The ratio of the respective heights attained by A and B after collision will be (neglect friction)

1 : 4

2 : 5

2 : 1

4 : 13

Since collision is elastic hence velocities will be exchanged. After collision vA​=2gh​ vB​=8gh​ HA​=2gvA2​​=h HB​=2gvB2​​=4h HB​HA​​=41​

Since collision is elastic hence velocities will be exchanged.
After collision $v_{A} = 2 g h$
$v_{B} = 8 g h$
$H_{A} = \frac{v _{A}^{2}}{2 g} = h$
$H_{B} = \frac{v _{B}^{2}}{2 g} = 4 h$
$\frac{H _{A}}{H _{B}} = \frac{1}{4}$