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.

Answer

A

Explanation

Force acting on the particle is constant. Hence, acceleration of the particle will also remain constant.
`a=F/m=2/1=2m//s^2`
Since, acceleration is constant. We can apply
`v=u+at`
and `s=ut+1/2at^2`
Refer Fig. 6.17 (a) `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)`
`2(sqrt2) m//s`
`alpha=tan^-1(|at|/|u|)=tan^-1(2/2) `
`=tan^-1(1)=45^@`

Thus, velocity of the particle at the end of `1s` is `2sqrt2 m//s` at an angle of `45^@` with its initial velocity.
Refer Fig.(b) `s=ut+1/2at^2`
Here, `ut` and `1/2at^2` are also two mutually perpendicular vectors. So,
`|s|=sqrt((|ut|)^2+(|1/2at^2|)^2)`
`=sqrt((2)^2+(1)^2)`
`sqrt5 m`
and `beta=tan^-1(|1/2at^2|/|ut|)`
`= tan^-1(1/2)`
Thus, displacement of the particle at the end of `1s` is `sqrt5 m` at an angle of `tan^-1(1/2)` from its initial velocity.