Problem 60 - Master Resource Book in JEE Main Physics

Question

A ball is dropped on the floor from a height of \(10 \mathrm{~m}\). It rebounds to a height of \(2.5 \mathrm{~m}\). If the ball is in contact with the floor for \(0.01 \mathrm{~s}\), the average acceleration during contact is nearly (Take \(g=10 \mathrm{~ms}^{-2}\) ) \(\begin{array}{ll}\text { (a) } 500 \sqrt{2} \mathrm{~ms}^{-2} \text { upwards } & \text { (b) } 1800 \mathrm{~ms}^{-2} \text { downwards }\end{array}\) \(\begin{array}{ll}\text { (c) } 1500 \sqrt{5} \mathrm{~ms}^{-2} \text { upwards } & \text { (d) } 1500 \sqrt{2} \mathrm{~ms}^{-2} \text { downwards }\end{array}\)

Step-by-Step Solution

Answer

The average acceleration is \(1500 \sqrt{2} \, \text{ms}^{-2}\) upwards.

Step 1: Determine initial velocity at impact

The ball is dropped from a height of 10m, so its initial velocity (u) before hitting the floor is 0. We can find the velocity it has when it hits the floor using the equation of motion: \( v^2 = u^2 + 2gh \), where \( g = 10 \, \text{ms}^{-2} \) and \( h = 10 \, \text{m} \). This gives \( v = \sqrt{0 + 2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \, \text{ms}^{-1} \).

Step 2: Determine velocity after rebound

We calculate how fast the ball rebounds by considering it rises to a height of 2.5 m. Using conservation of energy or kinematic equations, \( v^2 = u^2 - 2gh \). Thus, \( v^2 = 0 + 2 \times 10 \times 2.5 = 50 \), which means \( v = \sqrt{50} = 5\sqrt{2} \, \text{ms}^{-1} \). Rebound velocity is directed upwards.

Step 3: Calculate change in velocity

When the ball hits the ground, its velocity changes from \( 10\sqrt{2} \, \text{ms}^{-1} \) downwards to \( 5\sqrt{2} \, \text{ms}^{-1} \) upwards. The change in velocity is \( \Delta v = 5\sqrt{2} - (-10\sqrt{2}) = 15\sqrt{2} \, \text{ms}^{-1} \).

Step 4: Determine average acceleration

Average acceleration is defined as the change in velocity divided by the time taken for that change. We have \( \Delta v = 15\sqrt{2} \, \text{ms}^{-1} \) and contact time \( \Delta t = 0.01 \, \text{s} \). Thus, the average acceleration is \( a = \frac{15\sqrt{2}}{0.01} = 1500\sqrt{2} \, \text{ms}^{-2} \), and it is directed upwards.

Key Concepts

Kinematic Equations

Kinematic equations are powerful tools in physics that allow us to analyze the motion of objects under constant acceleration. They describe the relationships between displacement, velocity, acceleration, and time. These equations help us solve problems by connecting various aspects of motion without needing to evaluate forces directly.

In the context of the ball dropping from a height, we used the equation \( v^2 = u^2 + 2gh \) to find the impact velocity of the ball. Here:

• \( v \) is the final velocity just before impact.
• \( u \) is the initial velocity, which is 0 since the ball is dropped, not thrown.
• \( g \) is the acceleration due to gravity (\( 10 \mathrm{~ms}^{-2} \) in this scenario).
• \( h \) is the height from which the ball fell.

By plugging in the known values, we calculated the ball's velocity just before it hit the ground. Similarly, to find the rebound velocity, we rearranged and used: \( v^2 = u^2 - 2gh \). Though it may seem complex, breaking it down step-by-step makes it manageable.

Rebound Velocity

When an object bounces back after hitting a surface, like a dropped ball, the speed at which it leaves the surface is its rebound velocity. This concept is essential for understanding how energy transformations occur during collisions.

In this exercise, we examined the upward motion of the ball as it rebounded. Given the height it reached after the bounce, we applied the kinematic equation to find the maximum height's initial velocity. This velocity, known as rebound velocity, helps us gauge how much kinetic energy the ball retained despite energy losses during the impact with the ground.

Interestingly, in ideal theoretical scenarios where there's no energy loss, the ball would rise back to the original drop height. However, due to factors like energy loss to sound and heat, it only reaches a fraction of the original height—here, rising only to 2.5m from its origin of 10m.

Change in Velocity

The change in velocity is a critical concept for understanding how quickly an object can speed up, slow down, or change direction. It is the difference between an object's final and initial velocities and is key in calculating acceleration.

In the case of the bouncing ball, we calculated the change in velocity at impact. To do this, we took the initial downward velocity before hitting the ground (\(10\sqrt{2} \mathrm{~ms}^{-1}\)) and the upward rebound velocity (\(5\sqrt{2} \mathrm{~ms}^{-1}\)). The total change is seen as a combination of both the reversal of velocity direction and the reduction in magnitude.

Thus, the change in velocity (\(\Delta v\)) during impact was \( 15\sqrt{2} \mathrm{~ms}^{-1}\), illustrating how swiftly the ball changed its speed and direction in such a brief interaction with the floor. Calculating this value was necessary for determining the ball's average acceleration during contact, as acceleration is defined as change in velocity over time:\( a = \frac{\Delta v}{\Delta t} \).