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Question

A 0.098 kg block slides down a frictionless track as shown (Assume no energy loss on the entire track)
The time taken by the block to move from A to B is

A
1g
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B
2g
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C
3g
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D
4g
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Solution

The correct option is B 2g

Applying conservation of mechanical energy between point D and A,
Initial Potential energy (P.E) + initial kinetic enegy (K.E) = final potential energy (P.E) + kinetic energy (K.E)
mghD+12mV2D=mghA+12mV2A
mg(3)+12m(0)2=mg(1)+12mV2A
Solving the above equation we get,
VA=2g
Now vertcal component of VA i.e VAy=VAsin 30=g
When the block leaves the track at A it will undergo projectile motion.
Time of flight for projectile from A to B is given by,
T=2VAsin θg
Since VAsin θ=g (θ=30)
T=2gg=2g

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