A lift is moving down with acceleration a. A man  in the lift drops a ball inside the lift. The acceleration  of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively:

  • Option 1)

    g,g

  • Option 2)

    g - a,g - a

  • Option 3)

    g - a,g

  • Option 4)

    a,g

 

Answers (1)

As we learnt in

Lift accelerating down with 'a' -

V= variable

a<g

mg-R=ma

R=m(g-a)

- wherein

Apparent weight  <  Actual weight

 

 

 

For the man standing in the lift, the acceleration of the ball.

\overrightarrow{a_{bm}}=\overrightarrow{a_{b}}-\overrightarrow{a_{m}}\ \; \Rightarrow\ \; \overrightarrow{a_{bm}}=g-a

For the man standing on the ground, the acceleration of the ball. 

\overrightarrow{a_{bm}}=\overrightarrow{a_{b}}-\overrightarrow{a_{m}}\ \; \Rightarrow\ \; \overrightarrow{a_{bm}}=g-0=g

Correct option is 3.


Option 1)

g,g

This is an incorrect option.

Option 2)

g - a,g - a

This is an incorrect option.

Option 3)

g - a,g

This is the correct option.

Option 4)

a,g

This is an incorrect option.

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