Question

A person in an elevator accelerating upwards with an acceleration of $ 2m{s^{ - 2}} $ , tosses a coin vertically upwards with a speed of $ 20m{s^{ - 1}} $ . After how much time will the coin fall back into his hand? (Take $ g = 10m{s^{ - 2}} $ )
(A) $ \dfrac{5}{3}s $
(B) $ \dfrac{3}{{10}}s $
(C) $ \dfrac{{10}}{3}s $
(D) $ \dfrac{3}{5}s $

Answer 1

Hint: Find the effective acceleration of the coin and use the equation for 1-D motion to solve for the time taken. The effective acceleration of the coin with respect to the elevator will be $ {\mathbf{A}} = {{\mathbf{a}}_{\mathbf{s}}} - {{\mathbf{a}}_{\mathbf{l}}} $ where, $ {{\mathbf{a}}_{\mathbf{s}}} $ is the acceleration of the body with respect the inertial frame(here it is $ g $ ), $ {{\mathbf{a}}_{\mathbf{l}}} $ is the acceleration of the non-inertial frame(here it is the elevator) with respect to the inertial frame , $ {\mathbf{A}} $ is the acceleration of the coin in the lift . Then use the formula $ s = ut + \dfrac{1}{2}a{t^2} $ to find the time taken by the coin.

Complete answer:
The person is in an accelerating lift that means the frame of reference is a non- inertial frame of reference. So, first we have to find the net acceleration working on the coin in that frame of reference.
Now, we know if a body is moving in a non-inertial frame then net force with respect to the inertial frame of reference is $ {{\mathbf{a}}_{\mathbf{s}}}{\mathbf{ = }}{{\mathbf{a}}_{\mathbf{l}}} + {\mathbf{A}} $ where, $ {{\mathbf{a}}_{\mathbf{s}}} $ is the acceleration of the body with respect the inertial frame, $ {{\mathbf{a}}_{\mathbf{l}}} $ is the acceleration of the non-inertial frame(here it is the elevator) with respect to the inertial frame, $ {\mathbf{A}} $ is the acceleration of the body in the non inertial frame of reference. So, acceleration of the body in the non-inertial frame of reference(here the lift) will be, $ {\mathbf{A}} = {{\mathbf{a}}_{\mathbf{s}}} - {{\mathbf{a}}_{\mathbf{l}}} $
Here, we have, the lift is moving upwards and the gravitational field of earth is in the downward direction. So, the coin will have a net acceleration downward with respect to the lift.

So, given, $ {{\mathbf{a}}_{\mathbf{s}}} = {\mathbf{g = - 10\hat km}}{{\mathbf{s}}^{{\mathbf{ - 2}}}} $ and acceleration of the lift $ {{\mathbf{a}}_{\mathbf{l}}}{\mathbf{ = 2\hat km}}{{\mathbf{s}}^{{\mathbf{ - 2}}}} $ .Hence, effective acceleration of the coin with respect to the lift is, $ {\mathbf{A}} = ({\mathbf{ - 10\hat k}} - {\mathbf{2\hat k}}){\mathbf{m}}{{\mathbf{s}}^{{\mathbf{ - 2}}}} $ $ = {\mathbf{ - 1}}2{\mathbf{\hat km}}{{\mathbf{s}}^{{\mathbf{ - 2}}}} $ . Therefore acceleration of the particle is $ 12m{s^{ - 2}} $ in the downward direction with respect to the elevator.
Now, we can use the formula for the displacement of a particle moving in a straight line, $ s = ut + \dfrac{1}{2}a{t^2} $ Where, $ s $ is the displacement of the particle in time $ t $ , $ u $ is the initial velocity of the particle.
Here, we have given that, initial velocity of the particle is $ u = 20m{s^{ - 1}} $ and we calculated the acceleration of the coin as $ a = A = - 12m{s^{ - 2}} $ . The coin goes up and comes back to the hand of the person in the elevator hence, displacement of the coin here is zero, $ s = 0 $ .So, we have to solve for $ t $ .
Putting these values in the equation we get,
 $ 0 = 20t - \dfrac{1}{2} \cdot 12{t^2} $
 $ \Rightarrow 6{t^2} = 20t $
 $ \Rightarrow 3{t^2} = 10t $
Therefore, $ t $ becomes,
 $ \Rightarrow t = \dfrac{{10}}{3} $ [since, $ t \ne 0 $ ]
Therefore, the total time taken by the coin to go up and fall back to the hand is $ \dfrac{{10}}{3}s $ .
Hence, option (C ) is correct.

Note:
We can also solve the problem with respect to the frame of reference of earth .Then we have to consider that the lift is also going up and travels some distance upwards by the time the coin comes back. You can try that method on your own.