Problem 65 - Master Resource Book in JEE Main Physics

Question

A very broad elevator is going up vertically with a constant acceleration of \(2 \mathrm{~ms}^{-2}\). At the instant when its velocity is \(4 \mathrm{~ms}^{-1}\) a ball is projected from the floor of the list with a speed of \(4 \mathrm{~ms}^{-1}\) relative to the floor at an elevation of \(30^{\circ}\). The time taken by the ball to return the floor is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(1 / 2 s\) (b) \(1 / 3 \mathrm{~s}\) (c) \(1 / 4 \mathrm{~s}\) (d) \(1 \mathrm{~s}\)

Step-by-Step Solution

Answer

The time taken by the ball to return to the elevator floor is \( \frac{1}{2} \mathrm{~s} \).

Step 1: Identify the Problem

We need to determine the time it takes for a ball projected from a moving elevator to return to the elevator floor.

Step 2: Determine Initial Velocity Components

The ball is projected at an angle of 30° with respect to the horizontal from the elevator floor moving upwards. Thus, its initial velocity in the horizontal direction is \( u_x = 4 \cos(30) \) and in the vertical direction is \( u_y = 4 \sin(30) \).

Step 3: Break Down in Components

Calculate the initial velocities: \( u_x = 4 \cos(30) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \mathrm{~ms}^{-1} \) and \( u_y = 4 \sin(30) = 4 \times \frac{1}{2} = 2 \mathrm{~ms}^{-1} \).

Step 4: Calculate Net Vertical Acceleration

The effective acceleration acting on the ball vertically within the elevator is \( g - a = 10 \mathrm{~ms}^{-2} - 2 \mathrm{~ms}^{-2} = 8 \mathrm{~ms}^{-2} \) (since the elevator is accelerating upward).

Step 5: Set Up Vertical Motion Equation

Using the equation for vertical motion under constant acceleration: \( s = u_yt - \frac{1}{2}(g-a)t^2 \). We want the ball to return to the start position, so \( s = 0 \).

Step 6: Solve for Time

Substitute the values into the equation: \( 0 = 2t - \frac{1}{2} \times 8 \times t^2 \). Simplify to get: \( 0 = 2t - 4t^2 \). Factor the equation: \( t(2 - 4t) = 0 \). So, \( t = 0 \) or \( 2 - 4t = 0 \). Solve for \( t \): \( 4t = 2 \), \( t = \frac{1}{2} \).

Step 7: Conclude the Result

The time for the ball to return to the floor of the elevator is \( t = \frac{1}{2} \mathrm{~s} \).

Key Concepts

Projectile Motion

Projectile motion is a fascinating concept in physics where an object is launched into the air and moves along a curved path under the influence of gravity. This motion can be decomposed into two independent components: horizontal and vertical.

• In the horizontal direction, the object moves at a constant speed. This is because there are no external forces (like air resistance in ideal conditions) acting on it horizontally.
• Vertically, the object accelerates due to gravity— which means its speed changes over time.

In our exercise, a ball is projected from an elevator's floor at an angle of 30° from the horizontal. This means it has both vertical and horizontal velocity components.

Understanding these components is crucial. The horizontal motion doesn’t affect the vertical motion, considering ideal projectile motion without air resistance. Instead, both are affected by time but remain independent of each other in terms of their computation.

Vertical Acceleration

Vertical acceleration plays a key role in analyzing the projectile motion of the ball in our exercise. Normally, when objects are free-falling, the only vertical force acting on them is gravity, which provides an acceleration of approximately 9.8 \( \text{ms}^{-2} \). However, in this exercise, the ball is inside an accelerating elevator.

• The elevator exerts an additional upward acceleration of 2 \( \text{ms}^{-2} \).
• This reduces the effective gravitational acceleration acting on the ball. Instead of 10 \( \text{ms}^{-2} \), due to the lift's acceleration, the ball experiences an effective acceleration of 8 \( \text{ms}^{-2} \). This is obtained by subtracting the lift's acceleration from the gravitational acceleration: \( g - a \).

The effective acceleration is crucial to determining how fast the ball will ascend and descend during its flight. It contributes to how we set up our kinematic equations, which help calculate the time for the ball to return to the elevator floor.

Kinematic Equations

Kinematic equations are the backbone of solving projectile motion problems. They allow us to connect different aspects of motion, like displacement, initial velocity, time, and acceleration.

One useful equation is:

• \( s = u_yt - \frac{1}{2}(g - a)t^2 \) — where \( s \) is the displacement, \( u_y \) is the initial vertical velocity, and \( (g - a) \) is the effective downward acceleration.

These equations assume constant acceleration, which is key for simplifying our calculations. In the exercise, since we set the displacement \( s \) to zero (the ball returning to the floor where it started), it allowed us to solve for time using these kinematic principles.

By substituting known values, like \( u_y = 2 \ \text{ms}^{-1} \) and effective acceleration \( g - a = 8 \ \text{ms}^{-2} \), we can decipher the time \( t \) it takes for the ball to complete its journey. Using kinematic equations effectively helps solve complex motions with simpler calculations.