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Question

Two balls of equal masses are thrown upwards along the same vertical line at an interval of 2 seconds with the same initial velocity of 39.2ms1. The total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic, will be

A
5s and 3s
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B
10s and 6s
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C
515s and 315s
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D
(5+15)s and (3+15)s
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Solution

The correct option is D (5+15)s and (3+15)s
When the particles collide the height of both particles will be same,
For first particle time will be t and for second particle time will be t2
h=ut12gt2=u(t2)12g(t2)2

ut12gt2=ut2u12g(t24t+4)

0=2u+2gt2g

t=ug+1

t=5s

By solving the quadratic equation in t we get the time of first collision to be 5s.

h=39.2×512g×52=20g12.5g=7.5g meter

Now by momentum conservation, velocity of particles after collision
m(ugt)+m(ug(t2))=mv

mu5mg+mu3mg=mv

v=2u8g=0m/s

a=g

Now for collision with ground let time taken be t2

h=12gt22t=2hg=2×7.5gg=15

Thus, times of total time of flights will be 5+15s and 3+15s

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