Question

A car of mass M is moving with a uniform velocity v on a horizontal road when the hero of a
Hindi film drops himself on it from above. Taking the mass of the hero to be m, what will be
the velocity of the car after the event?​

Answer 1

Solution :

Consider the car plus the hero as the system. In the horizontal direction, there is no external force. Since the hero has fallen vertically , so his initial horizontal momentum = 0

Initial horizontal momentum of the system = Mv towards right

Finally the hero sticks to the roof of the car, so they move with equal horizontal velocity say V. Final horizontal momentum of the system

$\implies\tt (M+n )V \\ \\ \implies\tt Mv =(M+m)V \\ \\ \implies\tt V = \frac{Mv}{M+n}$

Answer 2

Answer:

v=mv/m+n

Explanation:

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