Problem 45 - Master Resource Book in JEE Main Physics

Question

A ball which is at rest, is dropped from a height \(h\) metre. As it bounces off the floor its speed is \(80 \%\) of what it was just before touching the ground? The ball \(\begin{array}{ll}\text { will then rise to nearly a height } & \text { [BVP Engg. 2007] }\end{array}\) (a) \(0.94 \mathrm{~h}\) (b) \(0.80 h\) (c) \(0.75 h\) (d) \(0.64 h\)

Step-by-Step Solution

Answer

(d) 0.64h

Step 1: Understanding the Problem

We are given a situation where a ball is dropped from a height \( h \) and after bouncing, its speed is reduced to 80% of the speed just before impact. We need to find out the height to which the ball will rise after the bounce.

Step 2: Determine Initial Conditions

Initially, the ball is at rest and is dropped from height \( h \). Using the formula for gravitational potential energy and kinetic energy, the ball's speed just before hitting the ground can be given by the formula: \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity.

Step 3: Determine Speed After Bounce

As the ball bounces, its speed is reduced to 80% of its previous speed. Therefore, the speed after bouncing is \( v' = 0.8v = 0.8 \times \sqrt{2gh} \).

Step 4: Calculate Maximum Height After Bounce

Apply the conservation of energy principle: the kinetic energy after bounce \( \left(\frac{1}{2}m(v')^2\right) \) is converted to potential energy at the maximum height \( \left(mgh'\right)\). This implies \( \frac{1}{2}m(0.8\sqrt{2gh})^2 = mgh' \). Simplifying this, we get: \( h' = 0.8^2 \times h = 0.64h \).

Step 5: Conclusion

The ball will rise to a height that is 64% of the original height \( h \). Therefore, the correct option is (d) 0.64h.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. When a ball is in motion just before hitting the ground, it has kinetic energy. This energy is calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the ball, and \( v \) is its velocity.

In the exercise, as the ball is dropped from a height, it starts accelerating due to gravity, converting its gravitational potential energy into kinetic energy. By the time it reaches the ground, all the potential energy converted to kinetic energy gives the ball speed. After bouncing, even though its speed reduces, the kinetic energy it retains will help the ball reach a new height.

• Initial kinetic energy just before the ball hits the ground is \( KE = \frac{1}{2}m(\sqrt{2gh})^2 \).
• After the bounce, kinetic energy becomes \( KE' = \frac{1}{2}m(v')^2 \), where \( v' = 0.8 \cdot \sqrt{2gh} \).

This kinetic energy transformation is crucial to figuring out how high the ball can bounce back up.

Conservation of Energy

The principle of conservation of energy states that energy can neither be created nor destroyed, only transformed from one form to another. In the context of our exercise, the total mechanical energy (potential energy + kinetic energy) of the ball is conserved if we ignore air resistance and other external forces.

Let's dive into how this works in our setup: when the ball initially drops from height \( h \), it starts with gravitational potential energy \( PE = mgh \) and essentially no kinetic energy. As it descends, this potential energy is continually converted into kinetic energy.

• At the peak height, all energy is potential.
• Just before hitting the ground, all energy is kinetic.

Upon impact and bounce, energy converts yet again as kinetic energy is partially converted back into potential energy, allowing the ball to rise. However, due to some energy loss (in this case, a reduction to 80% speed post-bounce), the ball doesn’t reach the original height. Hence, the potential energy at the peak of the bounce is less than it was originally. This translates to the height calculation: the final height post-bounce is \( h' = 0.64h \).

Acceleration due to Gravity

Acceleration due to gravity is a fundamental concept crucial to understanding motion under Earth's gravitational influence. Denoted by \( g \), it typically has a value of approximately \( 9.8 \text{ m/s}^2 \) on Earth's surface. This acceleration impacts the velocity of the ball as it falls from its original height.

In our scenario, as the ball drops, it accelerates downward because of gravity, gaining speed constantly until it hits the ground. This acceleration converts the gravitational potential energy into kinetic energy. When you calculate the velocity using the formula \( v = \sqrt{2gh} \), \( g \) plays a key role.

• The higher the value of \( g \), the faster the ball will accelerate downward.
• Gravity's constant action ensures that, initially at rest, the ball gains speed over time.

The bounce height eventually depends on the energy converted during descent and the retained kinetic energy post-bounce. Since gravity does not change, any loss in height on rebound is only due to energy loss through other means.