Extreme Momentum

At time t = 0 t=0 , a stationary particle of mass m m starts to experience a time dependent force F ( t ) = { k t ( T t ) for 0 t T 0 for t > T F(t) = \begin{cases} \mathbf{k}t(T-t) & \text{for } 0 \leq t \leq T \\ 0 & \text{for } t \gt T \end{cases}

where k \mathbf{k} is a constant vector.

Find the momentum of the particle when the action of the force discontinues.

  • If the answer is in the form k T α β \displaystyle \frac { \mathbf{k}{ T }^{ \alpha } }{ \beta } , where α \alpha and β \beta are natural numbers, find α + β \alpha + \beta .


The answer is 9.

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1 solution

Mv = f d t \int {f}{dt} integrating from 0 to T T we get the answer as α \alpha as 3 and β \beta as 6 so, 6+3 = 9 answer !

@Rishabh Cool bro were you able to get through JEE in 2016 ?

A Former Brilliant Member - 4 years, 5 months ago

Thanks for posting.

Swapnil Das - 4 years, 5 months ago

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oh ! no problem !

A Former Brilliant Member - 4 years, 5 months ago

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