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

a) mg b) 2mg c) $\large\frac{2}{3}$ mg d) $\large\frac{m^2g}{h}$

By conservation of energy
At A and B
$mg(3h)=mg(2h)+\large\frac{1}{2}$$mv^2 When v=velocity at B gh=\large\frac{1}{2}$$v^2$
$v=\sqrt {2gh}$
Also at B for the body to stay on the track, $N+mg=\large\frac{mv^2}{h}$
Substituting for $v^2$
$N+mg =2 mg$
$N=mg$
Hence a is the correct answer.
edited Jun 1, 2014