wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two balls having masses m and 2m are fastened to two light strings of same length l. The other ends of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocity of the balls just after their collision. (b) How high will the ball rise after the collision?
Figure

Open in App
Solution

Given:
Mass of the 1st ball = m
Mass of the 2nd ball = 2m

When the balls reach the lower end,
Let the velocity of m be v; and the velocity of 2m be v'.

Using the work-energy theorem, we can write:
12×mv2-12×m(0)2 = mgl v = 2glSimilarly, velocity of block having mass 2m will be,v' = 2gl v=v'=u (say)

Let the velocities of m and 2m after the collision be v1 and v2 respectively.

Using the law of conservation of momentum, we can write:
m×u-2mu=mv1+2mv2 v1+2v2 = -u ...(1)On applying collision formula, we get: v1-v2 = [(u-v)]=-2u ...(2)Substracting equation 1 from 2, we get:3v2 = u v2 = u3 = 2gl3Substituting this value in equation (1), we get:v1-v2 = -2u v1= -2u+v2 = -2u+u3 = -53u=-53×2gl = -50gl3

(b) Let the heights reached by balls 2m and m be h1 and h respectively.

Using the work energy principle, we get:
12×2m×(0)2 - 12×2m×v22 = -2m×g×h1 h1=l9Similarly,12×m×(0)2 - 12×mv12 = m×g×h212×50gl9 = g×h2h2 = 25l9

As h2 is more than 2l, the velocity at the highest point will not be zero.
Also, the mass m will rise by a distance 2l.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
A Sticky Situation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon