Understanding the Net Force of a Bouncing Ball on a Floor

This problem involves calculating the net force exerted by a ball on the floor during a bounce. To solve this, we need to determine the change in the ball's momentum during the collision and relate it to the impulse and the average net force acting on the ball. By Newton's third law, the net force exerted by the ball on the floor is equal in magnitude and opposite in direction to the net force exerted by the floor on the ball.

Given Information:

• Mass of the ball, \(m = 0.1\) kg
• Initial height, \(h_1 = 0.45\) m
• Rebound height, \(h_2 = 0.20\) m
• Contact time with the floor, \(\Delta t = 0.1\) s
• Gravitational acceleration, \(g = 10\) m s\(^{-2}\)

Calculating Velocity Before and After Collision

We can find the speed of the ball just before it hits the floor and just after it leaves the floor using the principle of conservation of energy (converting potential energy to kinetic energy). The speed \(v\) attained after falling from a height \(h\) or needed to reach a height \(h\) is given by the formula:

\(v = \sqrt{2gh}\)

Velocity before hitting the floor (\(v_1\)):

The ball falls from \(h_1 = 0.45\) m. Its speed just before impact is:

Let's take the downward direction as negative and the upward direction as positive. So, the velocity before impact is \(v_1 = -3\) m/s.

Velocity after leaving the floor (\(v_2\)):

The ball rises to \(h_2 = 0.20\) m. Its speed just after impact is:

Since the ball is moving upwards after impact, its velocity is \(v_2 = +2\) m/s.

Calculating Change in Momentum

Momentum is given by \(p = mv\). The change in momentum (\(\Delta p\)) of the ball during the collision is the final momentum minus the initial momentum.

\(\Delta p = p_{final} - p_{initial}\)

\(\Delta p = mv_2 - mv_1\)

Using the velocities with their signs (\(v_1 = -3\) m/s, \(v_2 = +2\) m/s) and mass \(m = 0.1\) kg:

The change in momentum is 0.5 kg m/s upwards.

Calculating Average Net Force on the Ball

The impulse-momentum theorem states that the impulse (force multiplied by the time over which it acts) is equal to the change in momentum. The impulse delivered by the floor to the ball is equal to the change in momentum of the ball.

\(Impulse = F_{net\_on\_ball} \times \Delta t = \Delta p\)

Here, \(F_{net\_on\_ball}\) is the average net force acting on the ball during the contact time \(\Delta t\).

This average net force of 5 N acts upwards on the ball. This net force is the vector sum of the force from the floor (\(F_{floor}\), upwards) and the gravitational force (\(mg\), downwards).

\(F_{net\_on\_ball} = F_{floor} - mg\)

\(5 \text{ N} = F_{floor} - (0.1 \text{ kg} \times 10 \text{ m/s}^2)\)

\(5 \text{ N} = F_{floor} - 1 \text{ N}\)

\(F_{floor} = 5 \text{ N} + 1 \text{ N}\)

\(F_{floor} = 6 \text{ N}\)

So, the average force exerted by the floor on the ball is 6 N upwards.

Net Force Applied by the Ball on the Floor

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The force exerted by the ball on the floor is the reaction force to the force exerted by the floor on the ball. The question asks for the *net* force the ball applied on the floor. This refers to the reaction to the *net* force the floor applied to the ball (which caused the change in momentum). The net force on the ball during contact was calculated as 5 N upwards.

Therefore, the net force applied by the ball on the floor is 5 N downwards. The magnitude of this net force is 5 N.

Summary of Calculation Steps:

1. Calculate velocity before impact using \(v_1 = \sqrt{2gh_1}\).
2. Calculate velocity after impact using \(v_2 = \sqrt{2gh_2}\).
3. Calculate the change in momentum, \( \Delta p = m(v_2 - v_1) \).
4. Calculate the average net force on the ball using \( F_{net\_on\_ball} = \frac{\Delta p}{\Delta t} \).
5. The magnitude of the net force applied by the ball on the floor is equal to the magnitude of the net force applied by the floor on the ball.

Using \(v_1 = -3\) m/s and \(v_2 = +2\) m/s:

The magnitude of the net force is 5 N.

Comparing this result with the given options, the net force applied on the floor while bouncing is 5.0 N.

Revision Table: Key Concepts

Additional Information: Force during Collision

During the brief contact time with the floor, the force exerted by the floor on the ball is not constant. It typically starts from zero, increases to a maximum, and then decreases back to zero. The calculations above determine the *average* net force over this contact time. The force from the floor is much larger than the gravitational force during the collision, causing the large change in velocity. The total upward force exerted by the floor includes supporting the ball's weight and providing the impulse to change its momentum direction and magnitude.

The total average force from the floor on the ball was calculated as 6 N upwards. This force counteracts gravity (1 N downwards) and provides the necessary upward acceleration resulting in a 5 N net upward force. By Newton's Third Law, the total average force exerted by the ball on the floor is 6 N downwards. However, the question specifically asks for the *net* force applied by the ball on the floor. This phrasing usually refers to the reaction to the *net* force on the ball, which is 5 N downwards. This aligns with option 5.0 N.