A particle of mass 1 kg has a velocity of `2m//s`. A constant force of 2N acts on the particle for 1s in a direction perpendicular to its initial velocity. Find the velocity and displacement of the particle at the end of 1 s.

Explanation

Force acting on the particle is constant. Hence, acceleration
`a=(F)/(m)=(2)/(1)=2ms^(-2)`
Since, acceleration is consstant. We can apply
`v=u+at` and `s=ut+(1)/(2)at^(2)`
`v=u+at`

Here, u and a t are two mutually perpendicular vectors. So
`|v|=sqrt((|u|)^(2)+(|at|)^(2))=sqrt((2)^(2)+(2)^(2))=2sqrt(2)ms^(-1)`
`alpha = "tan"^(-1)(|at|)/(|u|)=tan^(-1)((2)/(2))=tan^(-1)(1)=45^(@)`
Thus, velocity of the particle of the particle at the end of 1s is `2sqrt(2)ms^(-)` at an angle of `45^(@)` with its initial velocity.
`s=ut+(1)/(2)at^(2)`
Here, u t and `(1)/(2)at^(2)` are also two mutually perpendicular vector. So,
`|s|=sqrt((|ut|)^(2)+(|(1)/(2)a t^(2)|)^(2))=sqrt((2)^(2)+(1)^(2))=sqrt(5)m`
and `beta = "tan"^(-1)(|(1//2)a t^(2)|)/(|ut|)=tan^(-1)((1)/(2))`
Thus, displacement of the particle at the end of 1 s is `sqrt(5) m` at an angle of `tan^(-1)((1)/(2))` from initial velocity.