JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 15)
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. Here k is a constant. The value of T is :
$$2\sqrt {{k \over p}} $$
$$2\sqrt {{p \over k}} $$
$$\sqrt {{{2p} \over 2}} $$
$$\sqrt {{{2k} \over p}} $$
Explanation
$${{dp} \over {dt}} = F = kt$$
$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$
$$2p = {{K{T^2}} \over 2}$$
$$T = 2\sqrt {{P \over K}} $$
$$\int_P^{3P} {dP} = \int_0^T {kt\,dt} $$
$$2p = {{K{T^2}} \over 2}$$
$$T = 2\sqrt {{P \over K}} $$
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