A body of mass 1 kg is moving with velocity `30 ms^(-1)` due north. It is acted on by a force of 10 N due east for 4 seconds. Find the velocity of the body after the force ceases to act.

Explanation

M = 1kg, F=10 N due east , t=4s, u=0
`a=F/M =10/1 = 10 ms^(-2)` due east.
Final velocity due east at the end of 4 seconds.
v=u + at =0 + 10(4) = `40 ms^(-1)`
The body has initial velocity `30 ms^(-1)` due north. Which is not affected by applied force. At the end of 4 seconds. its velocity is `(40 hati + 30 hatj)`
`V= sqrt((40)^(2) + (30)^(2)) = 50 ms^(-1)`
Direction of resultant velocity `= tan alpha =30/40 = 3/4`
The resultant velocity makes an angle `alpha = tan^(-1) (3/4)` North east.