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Question

Two ball having masses m and 2m are fastened to two light strings of same length l (figure 9-E18). 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 velocities of the balls just after their collision. (b) How high will the ball rise after the collision?
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Solution

Let final velocity of m and 2m be v and v respectively.

Now according to law of conservation of momentum ,

m×X1+2m×V2=mV1+2mV2

m×u2mu=mV1+2mV2

V1=2V2=u................(1)

now,

V1V2=(V1V2)

V1V2=(uu)

V1V2=2u........................(2)

Subtracting Equation 1 and Equation 2

3V2=U

V2=U3=2gl3

Substituting value in Equation 2

V1V2=2U

V1=2U+V2

V1=2U+U3

V1=52gl3

V1=50gl3

(b) Again using Work Energy Principle

12×2mg212×V22=2mghh= height of heavy ball.

12×(2g9)=l×h

h=l9

Similarly ,

12×mg212mV21=mgh1 h_1 = height of light or small ball

h=50l18

Strings are fixed at O , hence maximum height of small ball of mass m ball will be 2l.


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