Understanding Acceleration in a Moving Lift

This question asks us to determine the acceleration of a ball dropped inside a lift that is moving downwards with a constant acceleration 'a'. We need to find this acceleration from two different viewpoints: one from a man in the lift and another from a man standing stationary on the ground. This involves understanding inertial and non-inertial frames of reference and how motion is observed in each.

Perspective 1: The Man on the Ground (Inertial Frame)

The man standing on the ground is in what we call an inertial frame of reference. This is a frame where Newton's laws of motion hold true without the need for fictitious forces.

• When the ball is dropped, the only significant force acting on it is the force of gravity exerted by the Earth.
• This force causes the ball to accelerate downwards.
• The acceleration due to gravity is denoted by 'g'.
• From the perspective of the man on the ground, the ball is simply falling under the influence of gravity, regardless of how the lift is moving.

Therefore, the acceleration of the ball as observed by the man on the ground is $\vec{g}$ downwards. Let's take downwards as the positive direction for acceleration. So, the acceleration is g.

Perspective 2: The Man in the Lift (Non-Inertial Frame)

The man in the lift is in a non-inertial frame of reference because the lift itself is accelerating. Observing motion from an accelerating frame can be a bit different.

• The lift is moving downwards with an acceleration 'a'.
• When the man drops the ball, initially the ball has the same velocity as the lift and the man.
• However, once dropped, the ball is under the influence of gravity (acceleration $\vec{g}$ downwards relative to the ground).
• To find the acceleration of the ball relative to the man in the lift, we can use the concept of relative acceleration between frames.

Let:

• $\vec{a}_{ball, ground}$ be the acceleration of the ball relative to the ground (inertial frame). This is $\vec{g}$ downwards.
• $\vec{a}_{lift, ground}$ be the acceleration of the lift relative to the ground. This is $\vec{a}$ downwards.
• $\vec{a}_{ball, lift}$ be the acceleration of the ball relative to the lift (what the man in the lift observes).

The relationship between these accelerations is given by:

$\vec{a}_{ball, ground} = \vec{a}_{ball, lift} + \vec{a}_{lift, ground}$

Let's again take downwards as the positive direction.

$g = \vec{a}_{ball, lift} + a$

Rearranging to find the acceleration of the ball relative to the lift:

$\vec{a}_{ball, lift} = g - a$

This means the acceleration of the ball as observed by the man in the lift is (g - a) downwards.

Summary of Accelerations

Based on our analysis of the ball's motion from the two different reference frames:

• Acceleration of the ball observed by the man in the lift: g - a (downwards)
• Acceleration of the ball observed by the man on the ground: g (downwards)

Therefore, the accelerations are (g - a, g) respectively.