Exams › JEE Main › Physics

A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. The value of T is:

1. 2√(k/p)
2. 2√(p/k)
3. √(2p/2)
4. √(2k/p)

Correct answer: 2√(p/k)

Solution

dp/dt = kt, so integral from 0 to T gives kT^2/2 = 3p - p = 2p. Thus T^2 = 4p/k and T = 2 sqrt(p/k).

Related JEE Main Physics questions

• Which of the following has the same SI unit as impulse?
• A monkey slides downward from a tree branch with uniform acceleration. If the maximum tension the branch can withstand is 75% of the monkey’s weight, what is the least downward acceleration that will allow the monkey to descend without snapping the branch?
• A 1000 kg car is travelling at 30 m/s. The brakes are applied until it stops. If the friction force between the tyres and the road is 5000 N, how long will the car take to come to rest?
• A rocket initially ejects gas from its rear at a mass rate of 0.1 kg s⁻¹. If the exhaust speed relative to the rocket is 50 m s⁻¹ and the rocket’s mass is 2 kg, what is its acceleration in m s⁻²?
• A block of mass M rests on a rough level floor with coefficient of friction μ. A person applies a horizontal pull, but the block remains at rest. If the total contact force exerted by the floor on the block is F, then
• Which of the following types of motion on a frictionless plane surface occurs without any force acting on the body?

⚔️ Practice JEE Main Physics free + battle 1v1 →