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 = 1 kg, 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 effected by applied force. At the end of 4 seconds, its velocity is `(40hat(i)+30hat(j))`.
`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 of ast.