Question

In the figure shown, a particle is released from the position A on a smooth track. When the particle reaches at B, then normal reaction on it by the track is:-

a) mg

b) 2mg

c) 2mg/3

d) m2g/h

Answer 1

Answer:

Option (a) mg

Explanation:

The figure is attached

When the particle reaches at B from A, the change in Potential Energy will be equal to Kinetic Energy

If the mass of the particle is m and velocity at the bottom most point is v then

$mg\times 3h-mg\times 2h=\frac{1}{2}mv^2$

$\implies v^2=2gh$

Since at B the particle is in circular motion, therefore a centripetal acceleration will be acting on it

The centripetal acceleration will be v²/h

At the point B if normal reaction is N then

$N+mg=m\frac{v^2}{h}$

$\implies N+mg=2mg$

$\implies N=mg$

Hope this helps,

Answer 2

Answer:

option a

Explanation:

velocity at B=✓u^2+2gh

=✓2gh

Now, circular motion equation

N+mg=m(v^2/R)

N+mg=m(2gh/u)

N=mg