Question

A 500 g block rests on a frictionless horizontal table at a distance of 400 nm from a fixed pin O. Then block is attached to pin O by an elastic cord of constant `k = 100 N//m` & of underformed length 900 mm. If the block is set in motion perpendiculariy as shown

A. the speed v to be set initially for which the distance from O to the block P will reach the maximum value of 1.2 m is 4.5 m/s
B. the speed u of the block at the moment when `OP = 1.2 m` is 1.5 m/s
C. the radius of curvature of the path of the block when `OP = 1.2 m` is 3.75 cm
D. A force of 30 N acts on the block along the length of the cord when `OP = 1.2 m`

Answer 1

Correct Answer - A::B::C::D
By conservation of angular momentum `0.5 xx V xx 0.4 = 0.5 xx u xx 1.2`
`v = 3u`
also by energy conservation

`(1)/(2) xx 0.5 xx v^(2) = (1)/(2) xx 100 xx (0.3)^(2) + (1)/(2) xx 0.5 xx u^(2)`
`rArr (v^(2))/(4) = (9)/(2) + (u^(2))/(4)`
`rArr (9u^(2))/(4) = (9)/(2) + (u^(2))/(4) rArr (8u^(2))/(4) = (9)/(2)`
`rArr u = sqrt(9)/(4) = (3)/(2)`
So `v = 3u = 3 xx 1.5 = 45`
(C ) `a_(n) = (u^(2))/(r) " " rArr r = (u^(2))/(a_(n))`
`(a_(n) = (K_(h))/(m) = (100 xx 0.3)/(0.5) = 60m//s^(2))`
`rArr r = ((1.5)^(2))/((60)) = (2.25)/(60) = 0.0375m`
`= 3.75cm`