The correct option is
D 1:25Given, just before collision, both the masses have same speed, let's say
v
Just After the collision, let the condition be as shown in the below figure.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/890363/original_02_S.png)
There is no external force present in the horizontal direction.
Hence, momentum will be conserved along horizontal direction.
⇒ 2mv−mv=2mv1+mv2
⇒ v=2v1+v2−−−−−−(1)
Collision is elastic hence,
e=1
⇒ e=speed of seperationspeed of approach=v2−v12v=1
⇒ 2v=v2−v1−−−−−(2)
from
(1) &
(2), we have
v2=5v3 ; v1=−v3
Here,
−ve represents that the assumed direction of
v1 is wrong, it will be opposite to the assumed direction.
So, the situation of the masses are as shown in the below figure.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/890365/original_02_S1.png)
Here, assumed height attained by the mass
2m is
h1 and by
m is
h2.
From Work-Energy theorm - (Just after collision to maximum height attained by the balls) we have,
For
2m mass ball
12(2m)v21+0=2mgh1+0
⇒ v1=√2gh1−−−−−(3)
Similarly, for
m mass ball
12mv22+0=mgh2+0
⇒ v2=√2gh2−−−−−−(4)
from eq
(3) &
(4), we have
v1v2=√2gh12gh2
⇒ ⎛⎜
⎜
⎜⎝v35v3⎞⎟
⎟
⎟⎠=√h1h2
⇒ (15)2=h1h2⇒ h1h2=125