Question

LI
CUCUR16
iv) A mass of 2m moving with some speed is directly approaching another mass m moving with
double speed. After some time, they collide with coefficient of restitution 0.5 . Ratio of
their respective speeds after collision is
a) 2/3 b) 3/2

d) ​

Answer 1

The ratio of  their respective speeds after collision is 1:2

Explanation:

• Let the speed of ball whose mass is 2m =  u1
• Before collision ball of mass m is at rest.
• Now after collision the speed of ball whose mass is 2m =  v1
• The speed of ball whose mass is m after collision = v2

 Now the coefficient of restitution e is given by "v1 - v2" = -e×u1 ..........(1)

Since it is given that v2 = u1

We can write equation.(1) as v1 - u1 = -e×u1   or v1 = (1-e)×u1 ................(2)

• By law of conservation of momentum  

2m×u1 = 2m×v1 + m×v2  = 2m×v1 + m×u1 ..............(3)

From eqn.(3), we get v1 = (1/2) u1   ..............(4)

From eqn.(2) and eqn.(4) we get,   e = 1/2

Thus the ratio of  their respective speeds after collision is 1:2

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