Question

A car of mass M is moving with uniform velocity v on a horizontal road. When a person of mass m drops on it from above, the velocity of the car will be?

• A. \( \frac{Mv}{M+m}\)
• B. \(\frac{Mv}{M}\)
• C. \(\frac{Mv}{m}\)
• D. \(\frac{Mv}{M+m}\)

Answer 1

The Correct Option is A

To solve this problem, we can use the principle of conservation of momentum.

Initially, only the car is moving, and the person is at rest. After the person drops onto the car, both are moving together.

Let's find the momentum before and after the person drops on the car:

Before:
Momentum of car = \(( M \times v )\)
Momentum of person = 0 (since the person is at rest with respect to the horizontal motion of the car) 
Total momentum before =\(( M \times v )\)

After: 
Let v' be the velocity of both the car and the person after the event. 
Total momentum after =\(( (M + m) \times v'\)) 

By the conservation of momentum, the total momentum before should be equal to the total momentum after: 
\([ M \times v = (M + m) \times v' ]\)

Rearranging to solve for v':
\([ v' = \frac{M \times v}{M + m} ]\)

So, the correct answer is: A \(( \frac{Mv}{M+m} )\)