Repeat above Question 3, if the charge is negative and the angle made by the boundary with the velocity is `theta = pi/6.`

Explanation

`2pi -2 theta =2pi-2xxpi/6=2pi-pi/3=(5pi)/3=omegat=(qB)/mt`
`implies t=(5pim)/(3qB)`
(b) Distance travelled `s=r(2pi-2 theta)=(5pir)/3`

(c) Impulse =charge in linear momentum
`=m(-v sin thetahati+vcostheta hatj)`
`-m(vsinthetahati+vcos thetahatj)`
`=-2mv sin thetahati=-2mvsin pi/6hati=-mvhatj`