Question

A force F = (2ti +3tj) N acts on an object moving in xy plane. Find magnitude of change in
momentum of the object in time interval t = 0 to t = 2 s.​

Answer 1

Answer:

kdowofjrjodorodocfefegtfvyfvtr53

Answer 2

Correct Question: A force F = (2ti +3t²j) N acts on an object moving in xy plane. Find magnitude of change in momentum of the object in time interval t = 0 to t = 2 s.​

Answer:

Momentum, |ΔP| = $\sqrt{80}$ N-s

Explanation:

Given;-

Force,  F = ( 2ti + 2t²j ) N _____(1)

We know that, area under curve of force-time graph relations gives change in momentum.

Now, we can write force as;-

            F = dP/ dt  ___(2) [f = rate of change in momentum w.r.t. time]

From (1) and (2), we have;-

          $\frac{dP}{dt}$ = 2ti + 2t²j

        $dP$ = $2t dt i + 3t^{2} dtj$

Integrating on both sides, we get;-

               $\int\limits^1_1 {dP} \, = 2 \int\limits^2_0 ({\frac{t^{2} }{2}) } \, + 3 \int\limits^2_0 ( {\frac{t^{3} }{3} )} \,$

                   ΔP = (2²)i + (2³)j

                   ΔP = 4i + 8j

                 |ΔP| = $\sqrt{4^{2}+ 8^{2} }$

                 |ΔP| = $\sqrt{16 + 64}$

                 |ΔP| = $\sqrt{80}$ N-s

Hence, the magnitude of change in momentum is $\sqrt{80}$ N-s.

Hope it helps! ;-))