Question

A body of mass 2 kg is moving along x-direction with velocity 2m/s. If a force of 4N is acting on it along y direction for 1 second, then what will be the final velocity

Answer 1

F=dpdt=m(v2−v1)t

4j^=2(v2−2i^)

2v2=4j+4i

|v2|=|2j+2i|

=22–√

Answer 2

Answer:

2$\sqrt{2$ m/s

Explanation:

Given:

Mass of the body, m = 2 kg

Force along the Y direction is

F=4j

Velocity along the X direction is

u=2i

Let v be the final velocity of the body after 1 s. Then, we have

Impulse=  change in linear momentum=  force x time

Ft= m(v-u)

4j x 1 = [2v-(2x2)i]

v=2i+2j

Magnitude of the final velocity is given by

v=$\sqrt{2^2+2^2$ m/s

v= 2$\sqrt{2}$ m/s

Note:

Here i , j are direction vectors and u, v, Fare vectors

That's a pretty good question !