Question : Two identical balls A and B are released from the position as shown in figure. They collide elastically on the horizontal portion. The ratio of heights attained by A and B after collision is (neglect friction)
a) 1:4
b) 2:1
c) 4:13
d) 2:5
Doubt by Jayant
Solution :
For Ball A
mg (4h) = ½m(v1)2
8gh v12
v1 = √(8gh)
Similarly for Ball B
v2 = √(2gh)
For Ball A
mg (4h) = ½m(v1)2
8gh v12
v1 = √(8gh)
Similarly for Ball B
v2 = √(2gh)
Since the two bodies are identical, i.e. m1 = m2 and the collision is identical.
Therefore, After the collision their velocities get interchanged.
So, v'1=v2 and v'2 = v1
Height gained by Ball A after collision
mgh1 = ½m(v'1)2
Therefore, After the collision their velocities get interchanged.
So, v'1=v2 and v'2 = v1
Height gained by Ball A after collision
mgh1 = ½m(v'1)2
mgh1 = ½m(v2)2
mgh1 = ½m (2gh)
h1 = h
h1 = h
Height gained by Ball B after collision
Let velocity of the ball after reaching the height h will be v.
Then
½m(v'2)2 = mgh + ½mv2
½m(8gh) = mgh + ½mv2
4mgh - mgh = ½mv2
3mgh = ½mv2
v = √(6gh)
Now Using Expression of Maximum Height from Projectile (Angular Projection Wala)
Hmax = v2 sin2θ/2g
Hmax = (6gh) × sin2 (60°) / 2g
Hmax = (6gh) × 3/4 × 1/2g
Hmax = 9h/4
Total Height Reached by the Ball B
h2 = h + Hmax
h2 = h + 9h/4
h2 = 13h/4
Required ratio = h1 : h2 = h : 13h/4
h1 : h2 = 4 : 13