Question

Problem-5.1 : A body of mass 1 kg is moving with
velocity 30 ms 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.​

Answer 1

Explanation:

Given,

Force act in direction of west F=10N

Velocity toward north Is V n =30 ms−1

Initial velocity in west direction is u=0m/s

Acceleration in west direction is a= M/F = 10/1

=10m/s ^2

Final velocity after time 4 sec, in west direction isV s =u+at=0+10×4=40m/s

Resultant velocity is V R = √V s^2 +V n^2 = √30^2 +40 ^2 =50ms ^−1

Direction of resultant velocity makes angle θ with the direction of force act.

tanθ= V s/V n = 40/30 = 4/3

θ=tan^ −1( 4/3 )

Hence the velocity of the body after the force ceases will be 50 m/s todards an angle θ=tan^−1( 4/3 ) with the direction of force.

i hope you understand