Problem 14 - College Physics

Question

A 2.0 -kg mud ball drops from rest at a height of \(15 \mathrm{~m}\). If the impact between the ball and the ground lasts \(0.50 \mathrm{~s}\), what is the average net force exerted by the ball on the ground?

Step-by-Step Solution

Answer

The average net force is 68.6 N upwards.

Step 1: Determine the Velocity Before Impact

To find the velocity of the mud ball just before it hits the ground, use the formula for gravitational acceleration: \(v = \sqrt{2gh}\). Here, \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity and \(h = 15\, \text{m}\) is the height. Plug the values into the equation: \(v = \sqrt{2 \times 9.8 \times 15}\). Calculate to get \(v \approx 17.15\, \text{m/s}\).

Step 2: Calculate the Change in Momentum

The change in momentum \((\Delta p)\) can be calculated using the formula: \(\Delta p = m \cdot \Delta v\), where \(m = 2.0\, \text{kg}\) is the mass of the mud ball, and \(\Delta v = -17.15 - 0\, \text{m/s}\) since the ball comes to rest. Thus, \(\Delta p = 2.0 \times (-17.15) = -34.3\, \text{kg m/s}\).

Step 3: Compute the Average Force

Use the impulse-momentum theorem \(F_{avg} \cdot \Delta t = \Delta p\) to find the average force. \(F_{avg}\) is the average force, and \(\Delta t = 0.5\, \text{s}\) is the duration of the impact. Rearrange to find \(F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-34.3}{0.5} = -68.6\, \text{N}\). The negative sign indicates the force is directed upwards, opposite to the mud ball's initial motion.

Key Concepts

Gravitational Acceleration

Gravitational acceleration is a constant that describes the increase in velocity of an object as it falls freely under the influence of gravity.
On Earth, this value is typically represented by \( g = 9.8 \, \text{m/s}^2 \). It plays a significant role in determining how fast an object will move as it descends. To calculate how fast an object will be moving just before it hits the ground, one can use the formula \( v = \sqrt{2gh} \).
In this scenario, the 2.0-kg mud ball is falling from a height of 15 meters. By applying the gravitational acceleration, we find that the velocity just before the ball impacts the ground is approximately 17.15 m/s. This measurement helps establish how fast gravity accelerates the mud ball during its free fall.

Impulse-Momentum Theorem

The impulse-momentum theorem connects the impulse exerted on an object to its change in momentum. This principle is articulated with the equation \( F_{avg} \cdot \Delta t = \Delta p \).
Here, \( F_{avg} \) represents the average force, and \( \Delta t \) signifies the time duration of the force application. This theorem is vital when analyzing collision scenarios where forces act over short time intervals.

• Impulse refers to the product of the average force and the time duration over which it acts.
• The change of momentum \( \Delta p \) represents the variation in the object's momentum due to the impulse.

In the mud ball example, during the 0.50 second impact, the average net force equates to the time-weighted change of momentum of the mud ball. This showcases how force and time interrelate to alter the momentum of a body.

Change in Momentum

Momentum is a measure of the quantity of motion an object possesses, calculated as the product of an object’s mass \( m \) and its velocity \( v \). Hence, the momentum \( p \) is expressed as \( p = m \cdot v \).
The change in momentum, \( \Delta p \), occurs when an object's velocity changes, resulting in a new momentum value. The formula \( \Delta p = m \cdot \Delta v \) expresses this shift. In this context, \( \Delta v \) is the change in velocity, which for the mud ball is \(-17.15 \,- 0 \), given that it goes from moving to rest.

• The initial momentum is calculated at the moment before impact using both the mass and the velocity derived.
• Post-collision, since the mud ball comes to rest, its final momentum is zero.

Therefore, the calculation of \( \Delta p = -34.3 \, \text{kg m/s} \) captures the complete stoppage of the mud ball's motion upon impact, illustrating the direct effect of external forces in stopping its movement.