Problem 56 - College Physics

Question

A steel ball with a mass of \(40.0 \mathrm{~g}\) is dropped from a height of \(2.00 \mathrm{~m}\) onto a horizontal steel slab. The ball rebounds to a height of \(1.60 \mathrm{~m}\). (a) Calculate the impulse delivered to the ball during the impact. (b) If the ball is in contact with the slab for \(2.00 \mathrm{~ms},\) find the magnitude and direction of the average force on the ball during the impact. (c) How much mechanical energy was lost in the impact with the steel slab? (d) How high would the ball have rebounded if the impact had been perfectly elastic?

Step-by-Step Solution

Answer

(a) Impulse is 0.4744 kg·m/s. (b) The average force is 237.2 N upward. (c) 0.1568 J of mechanical energy was lost. (d) The ball would rebound to 2.00 m if the impact were perfectly elastic.

Step 1: Calculate initial and final velocities

The initial velocity of the ball before impact (\( v_i \), downward) can be found using the formula: \[ v_i = \sqrt{2gh} \]where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity) and \( h = 2.00 \, \text{m} \). Substituting the values:\[ v_i = \sqrt{2 \times 9.8 \times 2.00} = \sqrt{39.2} \approx 6.26 \, \text{m/s} \]The final velocity of the ball after impact (\( v_f \), upward) when rebounding to 1.60 m is:\[ v_f = \sqrt{2gh_f} \]where \( h_f = 1.60 \, \text{m} \). Substituting the values:\[ v_f = \sqrt{2 \times 9.8 \times 1.60} = \sqrt{31.36} \approx 5.60 \, \text{m/s} \]

Step 2: Calculate the impulse delivered to the ball

Impulse is defined as the change in momentum. The ball's momentum change can be expressed as:\[ \text{Impulse} = \Delta p = m(v_f - (-v_i)) \]where the mass \( m = 40.0 \, \text{g} = 0.040 \text{ kg} \).\[ \Delta p = 0.040(5.60 + 6.26) = 0.040 \times 11.86 = 0.4744 \, \text{kg m/s} \]

Step 3: Calculate the average force on the ball

The average force \( F \) can be found using the formula:\[ F = \frac{\Delta p}{\Delta t} \]where \( \Delta t = 2.00 \, \text{ms} = 0.002 \, \text{s} \).\[ F = \frac{0.4744}{0.002} = 237.2 \, \text{N} \]The force is directed upwards as the ball rebounds.

Step 4: Calculate mechanical energy lost during the impact

The mechanical energy before the impact is:\[ E_i = mgh = 0.040 \times 9.8 \times 2.00 = 0.784 \, \text{J} \]The mechanical energy after the impact is:\[ E_f = mgh_f = 0.040 \times 9.8 \times 1.60 = 0.6272 \, \text{J} \]The energy lost during the impact is:\[ \Delta E = E_i - E_f = 0.784 - 0.6272 = 0.1568 \, \text{J} \]

Step 5: Calculate rebound height for a perfectly elastic collision

In a perfectly elastic collision, no kinetic energy is lost. Therefore, the total energy after the impact equals the initial energy,\[ E_f = E_i \]The rebound height \( h' \) can be calculated from:\[ mgh' = mgh \implies h' = h = 2.00 \, \text{m} \]

Key Concepts

Momentum Change

When the steel ball hits the slab and bounces back, its momentum changes. Momentum is calculated as the product of mass and velocity. The change in momentum, or impulse, occurs because the ball's velocity changes direction during the collision.

The initial momentum is computed right before the ball impacts the ground, using its mass and initial velocity. Similarly, the final momentum is calculated right after the rebound using the mass and final velocity. The impulse is the vector sum of these two momentum values. Here, \(\Delta p = m(v_f - (-v_i))\), where \(m\) is the mass of the ball, \(v_i\) is the velocity just before impact, and \(v_f\) is the velocity as it rebounds.

Understanding the momentum change is fundamental because it helps connect the instantaneous effect (impulse) with the forces applied over a period.

Average Force Calculation

The average force on an object is the total impulse exerted on it divided by the time duration of the impulse. In our scenario, the average force reflects the constant interaction between the ball and the slab during their brief contact period.

This average force can be found using the equation \( F = \frac{\Delta p}{\Delta t} \), where \(\Delta p\) is the impulse delivered to the ball and \(\Delta t\) is the contact time. For the ball, which was in contact with the slab for \(2.00 \text{ ms} = 0.002 \text{ s}\), we can compute the force by substituting these values. This results in a significant rise in force due to the short impact duration.

Grasping how to calculate this force allows us to understand how strong forces can arise from over very short times in collisions.

Mechanical Energy Loss

In many impacts, especially inelastic ones, mechanical energy is not conserved. In this exercise, mechanical energy is lost when the ball hits the slab and rebounds.

Initially, the potential energy of the ball converts entirely into kinetic energy as it falls. However, after the impact, some energy is dissipated as heat and sound, or remains as deformation energy in the slab and ball, leading to less energy being conserved as mechanical energy post-collision.

We calculate this by contrasting the mechanical energy before and after the impact. The initial energy is \(E_i = mgh \) and the final energy is \(E_f = mgh_f\). The lost energy \(\Delta E = E_i - E_f\) shows how much energy didn’t transform back into potential energy when the ball rebounded.

Elastic and Inelastic Collisions

Collisions are classified as either elastic or inelastic based on energy conservation. In elastic collisions, total kinetic energy is conserved, meaning the ball rebounds at the same height it was dropped from, thus losing no mechanical energy to other forms.

Conversely, in an inelastic collision like in this scenario, some of the kinetic energy is lost. We see this because the ball does not reach its original height upon rebounding, indicating energy transformed into other forms or maintained as internal deformation energy.

Solving the problem under the assumption of a perfectly elastic collision allows the ball to rise back to its initial height, maintaining \(E_f = E_i\). This hypothetical helps us understand real-world scenarios where factors like perfectly elastic collisions are rarely seen. It also highlights the importance of detailing interactions in mechanical energy studies.