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Question

A particle is thrown up with velocity v0 inside a stationary lift of sufficient height. The time of flight is 'T'. Now it is thrown again with same initial speed v0 with respect to lift. If at the time of second thrown, lift is moving up with a velocity v0 and uniform acceleration g upward. Which of the following are true?


A

(a)Displacement-time graph of particle when lift was stationary

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B

(b)Displacement-time graph of the particle with respect to the moving lift.

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C

(c)Displacement-time graph of the particle with respect to ground when the lift is moving

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D

(d) Velocity-time graph of the particle with respect to the moving lift.

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Solution

The correct options are
A

(a)Displacement-time graph of particle when lift was stationary


B

(b)Displacement-time graph of the particle with respect to the moving lift.


C

(c)Displacement-time graph of the particle with respect to ground when the lift is moving


D

(d) Velocity-time graph of the particle with respect to the moving lift.


When the lift was stationary:- velocity of projection = v0
Time of flight = T so after T particle comes back so displacement = 0
S=uT12gT2
t(v012gT)=02v0g=T
We know the velocity keeps decreasing, till it becomes 0. So s-t curve will be parabolic.
To find maximum height
v20=2gs s=v2o2g
So graph (a) is correct


When lift is moving:
Lift is moving up with v0, particle is thrown with velocity v0 relative to lift i.e. If i am the lift, I will see the particle leaving me with velocity v0vp/l=v0 ....(1)
Now accerleration of lift, a1=+g
Accerleration of particle =ap=g
ap/l=gg=2g ....(2)
Now same laws of motion are applied here:
s=v0+12(2g)t2 when particle comes back s=0
0=v0+gt2
0=t(v0gt)
t=0,v0g
So particle comes back in v0g or T2
From lifts frame, the maximum height will be attained by particle when its velocity w.r.t. lift = 0
0(v0)2=2(2g)s
s=v204g=S2

from lifts frame, the maximum height will be attained by particle when its velocity w.r.t. lift =0
0(v0)2=2(2g)s
s=v204g=S2


So graph s is correct position time graph of particle w.r.t. the lift.
Lift is moving but from grounds from of reference
We know that after being thrown, particle hits lift in T2 or v0g seconds
Also, particle was thrown at velocity v0 w.r.t. the lift
i.e. vp/l=v0
Velocity of lift=v0
vp/l=vpvl
v0=vpv0
ap=g downwards

Particle is caught by lift only after T/2. We need to find the displacement till that time

s=v0T212g(T2)2
v024v0g1b4v0g


s=3v202g=3s
So the graph c is correct


Now,
vp/l=v0
ap/l=2g
vp/l=up/l2gt
vp/l=v02gt


At, t=0
vp/l=v0
Slope is constant i.e. -2g
If vp/l=v0
Then, t2v02g=T4


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