Question

A force F(t) = 12 - 3t (N) varying with time acts on
a particle moving along x-axis. Starting from v = 0
particle is taken along x-axis. Find the momentum
of particle at t = 10 s.​

Answer 1

according to Newton's 2nd law, force is the rate of linear momentum with respect to time.

i.e., F = dP/dt

or, $\int\limits^{10}_0{F}\,dt=\int\limits^{P_f}_{P_i}{dP}$

given, force is a function of time such that F(t) = 12 - 3t

so, $\int\limits^{10}_0{(12-3t)}\,dt=P_f-P_i$

it is given that, Initial velocity is zero.

we know, linear momentum is mass × velocity. so, initial linear momentum = 0

now, $P_f=[12t-1.5t^2]^{10}_0$

= 12(10 - 0) - 1.5(10² - 0²)

= 120 - 225

= - 105 Kgm/s

hence, momentum at t = 10s is -105 Kgm/s.

Answer 2

Answer: -30 m/s

Explanation:

integration of P = integration of F dt

= integration (12-3t) dt

= 12t-3t^2/2

= 12(10) - 3(10)^2/2

= 120-300/2 = 120-150

= -30 m/s