Question

A particle is moving on a straight line x + y = 2.
Its angular momentum about origin is L = 3t + 2
(kg mềs-"). Find the force acting on the particle at
t = 2 s. (x and y are in metre)
(1)
v2
N.
(2) 3N
(3) 3V2N
(4) VZN​

Answer 1

$\sf \Large Solution!!$

We know that Angular Momentum = Linear Momentum × Lower arm

and we know that Lower arm = Perpendicular length = √2 ($\therefore$ x+ y = 2)

Now,

3t + 2 = mv × √2

diff. both side,

3 = $\sf \dfrac{mdv}{dt}\times \sqrt{2}$

Therefore, $\sf \dfrac{mdv}{dt}=F$

F = $\sf \dfrac{3}{\sqrt{2}}N$