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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 39.2ms−1. 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)sWhen 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 t−2h=ut−12gt2=u(t−2)−12g(t−2)2ut−12gt2=ut−2u−12g(t2−4t+4)0=−2u+2gt−2gt=ug+1t=5sBy solving the quadratic equation in t we get the time of first collision to be 5s.h=39.2×5−12g×52=20g−12.5g=7.5g meterNow by momentum conservation, velocity of particles after collisionm(u−gt)+m(u−g(t−2))=mv′mu−5mg+mu−3mg=mv′v′=2u−8g=0m/sa=−gNow for collision with ground let time taken be t2h=12gt22⇒t=√2hg=√2×7.5gg=√15Thus, times of total time of flights will be 5+√15s and 3+√15s

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