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Chapter 4: Problem 1

From a lift moving upwards with a uniform acceleration \(a=2\) \(\mathrm{ms}^{-2}\), a man throws a ball vertically upwards with a velocity \(v=12 \mathrm{~ms}^{-1}\) relative to the lift. The ball comes back to the man after a time \(t\). Find the value of \(t\) in seconds.

Short Answer

Expert verified
The time \( t \) is approximately 3.08 seconds.

Step by step solution

01

Determine the Effective Acceleration of the Ball

The ball is thrown from a lift that has an acceleration of \( 2 \ \text{ms}^{-2} \) upwards. Considering the gravitational acceleration \( g = 9.8 \ \text{ms}^{-2} \) (acting downwards), the effective acceleration \( a_e \) experienced by the ball becomes the difference between gravitational acceleration and the lift's acceleration. Thus, \( a_e = g - a = 9.8 \ \text{ms}^{-2} - 2 \ \text{ms}^{-2} = 7.8 \ \text{ms}^{-2} \) downward.
02

Use Kinematic Equation to Set Up the Problem

When a ball is thrown upwards with an initial velocity \( v = 12 \ \text{ms}^{-1} \) and returns to the starting point, the vertical displacement \( s = 0 \). We use the kinematic equation: \[ s = vt - \frac{1}{2}at^2 \]Substituting \( s = 0 \), we get:\[ 0 = 12t - \frac{1}{2} \times 7.8 \times t^2 \].
03

Solve the Quadratic Equation for Time

The equation simplifies to:\[ 0 = 12t - 3.9t^2 \]Factor out \( t \) from the equation:\[ t(12 - 3.9t) = 0 \]This gives two solutions: \( t = 0 \) or \( 12 - 3.9t = 0 \). Since \( t = 0 \) represents the initial time of projection (which we are not evaluating), focus on:\[ 12 - 3.9t = 0 \]Simplify to find \( t \):\[ 3.9t = 12 \]\[ t = \frac{12}{3.9} \approx 3.08 \text{ seconds} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations and their Applications
Kinematic equations are essential in classical mechanics because they describe the motion of objects under constant acceleration. They allow us to relate various parameters such as displacement, initial and final velocities, acceleration, and time.

Let's look at the key kinematic equation used in this context where the ball is projected:
  • Displacement: \( s = vt - \frac{1}{2}at^2 \)
For a ball thrown upward and coming back to the starting point, the displacement \( s \) is zero. This is because it rises to a maximum height and then returns.
  • When you substitute \( s=0 \), initial velocity \( v=12 \ \text{ms}^{-1}\), and effective acceleration as \( a=7.8 \ \text{ms}^{-2}\) in the equation, you can solve for time \( t \).
This equation simplifies to a quadratic form: \( 0 = 12t - 3.9t^2 \). Such quadratic equations may offer two possible solutions for time, and you must choose the one that makes sense physically鈥攈ere, it鈥檚 when the ball returns.
Understanding Effective Acceleration
Effective acceleration is a concept that arises when dealing with objects in a non-inertial frame of reference, such as a moving lift or elevator. In this problem, the lift moves upwards with a constant acceleration while gravity acts downward.

So, to find the "effective" acceleration experienced by the ball, you subtract the lift's acceleration from gravitational acceleration. Gravitational acceleration \( g \) is \( 9.8 \ \text{ms}^{-2}\) downward, and the lift's acceleration \( a = 2 \ \text{ms}^{-2}\) is upward. The effective acceleration of the ball \( a_e \) is computed as follows:
  • \( a_e = g - a = 9.8 - 2 = 7.8 \ \text{ms}^{-2} \) downward.
This acceleration accounts for the combined effect of gravity and the lift's motion on the ball, dictating how the ball moves relative to the lift.
Exploring Relative Motion
Relative motion describes how the position or velocity of an object is perceived differently when compared to another moving object or reference frame. In this scenario, the ball is thrown within a lift that itself is accelerating upwards.

The key to understanding relative motion is to recognize how the ball's initial velocity is described relative to the lift, not a stationary ground observer. The ball moves upwards with an initial velocity of \( 12 \ \text{ms}^{-1} \) relative to the lift. But, because the lift is accelerating upwards, any observer inside the lift sees a different motion pattern than an observer outside.

In essence, relative motion analysis lets us appreciate how velocities add up: the ball's velocity relative to the lift is affected by the motion of the lift itself. Thus, it's crucial to adjust our calculations based on this frame of reference to deduce the time it takes for the ball to return, considering the lift's movement.

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Most popular questions from this chapter

The displacement \(x\) of a particle moving in one dimension under the action of a constant force is related to time \(t\) by the equation \(t=\sqrt{x}+3\), where \(x\) is in meters and \(t\) is in seconds. Find the displacement of the particle when its velocity is zero. (1) Zero (2) \(12 \mathrm{~m}\) (3) \(6 \mathrm{~m}\) (4) \(18 \mathrm{~m}\)

A passenger reaches the platform and finds that the second least boggy of the train is passing him. The second last boggy takes 3 s to pass the passenger, and the last boggy takes \(2 \mathrm{~s}\) to pass him. Find the time by which the passenger late for the departure of the train? Assume that the train accelerates at constant rate and all the boggies are of equal length.

A point moves such that its displacement as a function of time is given by \(x^{3}=t^{3}+1\). Its acceleration as a function of time \(t\) will be (1) \(\frac{2}{x^{5}}\) (2) \(\frac{2 t}{x^{5}}\) (3) \(\frac{2 t}{x^{4}}\) (4) \(\frac{2 t^{2}}{x^{5}}\)

A steel ball is dropped from the roof of a building. A man standing in front of a \(1-m\) high window in the building notes that the ball takes \(0.1\) s to the fall from the top to the bottom of the window. The ball continues to fall and strikes the ground. On striking the ground, the ball gets rebounded with the same speed with which it hits the ground. If the ball reappears at the bottom of the window \(2 \mathrm{~s}\) after passing the bottom of the window on the way down, find the height of the building.

Two bodies of masses \(m_{1}\) and \(m_{2}\) are dropped from heights \(h_{1}\) and \(h_{2}\), respectively. They reach the ground after time \(t_{1}\) and \(t_{2}\) and strike the ground with \(v_{1}\) and \(v_{2}\), respectively. Choose the correct relations from the following. (1) \(\frac{t_{1}}{t_{2}}=\sqrt{\frac{h_{1}}{h_{2}}}\) (2) \(\frac{t_{1}}{t_{2}}=\sqrt{\frac{h_{2}}{h_{1}}}\) (3) \(\frac{v_{1}}{v_{2}}=\sqrt{\frac{h_{1}}{h_{2}}}\) (4) \(\frac{v_{1}}{v_{2}}=\frac{h_{2}}{h_{1}}\)
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