Question

A 2.0kg mass is intally moving at a speed of 10m/s due north, when a constant

eastward 10.ON force is applied to it for 3.0 seconds. Which direction is the mass going after that?
(A) 27° east of north
(B) 56. east of north
(C) 34. east of north
(D)18. east of north

Answer 1

Option(B) 56. east of north

Step-by-step explanation:

2D Motion

If objects are moving in an XY plane subject to net force (F), acceleration (a), velocity (v) and displacement (r) in both axes, we must consider all those magnitudes as vectors because they have components in x and y.

If
`\vec a` is the acceleration vector, then

`\vec v=\vec v_o+\vec a.t`

`\vec r=\vec v_o.t+(\vec a.t^2)/(2)+\vec r_o`

`\vec F=m.\vec a`

Assuming the positive direction to the right and upwards, we are given the following data

`m=2\ kg`

`\vec v_o=0\ m/s\ \hat i+10\ \hat j`

The x-component of the velocity is zero because it due north.

The force is applied eastward;

`\vec F=10\ \hat i+0\ N\ \hat j`

t=3 sec

From the formula

`\vec F=m.\vec a`

we can solve for
`\vec a`

`\displaystyle \vec a=(\vec F)/(m)`

`\displaystyle \vec a=(10\ \hat i+0\ N\ \hat j)/(2\ kg)`

`\displaystyle \vec a=5\ m/sec^2\ \hat i+0\ \hat j`

We can now compute the velocity at t=3 sec

`\vec v=0\ \hat i+10\ m/s\ \hat j+\ (5\ m/sec^2\ \hat i+0\ \hat j)3\ sec`

Adding and simplifying

`\vec v=15\ m/s\ \hat i+10\ m/s\ \ \hat j`

The direction is given by the angle computed as

`\displaystyle tan\theta=(v_y)/(v_x)`

`\displaystyle tan\theta=(10)/(15)`

`\theta=arctan(0.667)`

`\theta=33.7^o\approx 34^o`

This angle is north of east, the required angle is

`90^o-34^o=56^o`