Question

A small particle initially at point pstarts moving fromrest.The whole space where particle will move is divided into three regions as shown in figure.In region (i) particle accelerates through (5m/s^(2)) where direction of acceleration is along the normal of the screen while in region (ii) the acceleration acts in such a way that it is always perpendicular to the direction of motion resulting the particle to move on a circulat track having radius (20)/(sqrt(3))m .There is uniformacceleration in region (ii) in such a manner that velocity of particle become thrice (without change in direction) when it just reach the screen.Find the average speed of a particle (in m/s). qquad (1)/(4)

Answer 1

Answer:

Initial height is r(1−cosα)

Final height is r(1−sinβ)

Net vertical displacement is r(1−sinβ)−r(1−cosα)=r(cosα−sinβ)

Apply equation of kinematic

v

2

−u

2

=2as

v

2

=2gr(cosα−sinβ)

Normal reaction,

N=mgsinβ−

r

mv

2

0=mgsinβ−

r

m2gr(cosα−sinβ)

3sinβ=2cosα

Hence, the relation is 3sinβ=2cosα