Problem 67 - University Physics with Modern Physics

Question

A steel ball with mass 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 impact. (b) If the ball is in contact with the slab for 2.00 \(\mathrm{ms}\) , find the average force on the ball during impact.

Step-by-Step Solution

Answer

The impulse is approximately 0.474 Ns, and the average force is about 237 N.

Step 1: Convert Units of Mass

First, convert the mass of the ball from grams to kilograms. Since there are 1000 grams in a kilogram, the mass of the ball is given by:\[ m = 40.0\ \text{g} = 40.0 \times 10^{-3}\ \text{kg} = 0.040\ \text{kg} \]

Step 2: Calculate Initial Velocity of the Ball (Before Impact)

The ball is dropped from a height of 2.00 m. Use the kinematic equation for motion under gravity, ignoring air resistance, to find the velocity just before impact.Initial velocity \((v_i)\) is found by:\[ v_i = \sqrt{2gh_1} \]Where:- \(g = 9.8\ \text{m/s}^2\) is the acceleration due to gravity.- \(h_1 = 2.00\ \text{m}\)Calculate:\[ v_i = \sqrt{2 \times 9.8 \times 2.00} = \sqrt{39.2} \approx 6.26\ \text{m/s} \]

Step 3: Calculate Final Velocity of the Ball (After Rebound)

The ball rebounds to a height of 1.60 m. Use the kinematic equation to find the velocity just after leaving the slab.Final velocity \((v_f)\) is found by:\[ v_f = \sqrt{2gh_2} \]Where:- \(h_2 = 1.60\ \text{m}\)Calculate:\[ v_f = \sqrt{2 \times 9.8 \times 1.60} = \sqrt{31.36} \approx 5.60\ \text{m/s} \] Since the ball is rebounding upwards, this velocity is considered positive.

Step 4: Calculate Impulse Delivered to the Ball

Impulse is the change in momentum of the ball. Since momentum \(p = mv\), the impulse \(J\) is given by:\[ J = m(v_f - (-v_i)) = m(v_f + v_i) \]Substitute the known values:\[ J = 0.040\ \text{kg} \times (5.60\ \text{m/s} + 6.26\ \text{m/s}) \]Calculate:\[ J = 0.040 \times 11.86 = 0.4744\ \text{Ns} \approx 0.474\ \text{Ns} \]

Step 5: Calculate Average Force During Impact

Average force \(F_{avg}\) during the impact is calculated using the impulse-time relationship, where impulse \(J\) is the product of the average force and the time \(t\) of contact:\[ F_{avg} = \frac{J}{t} \]Given the time of contact, \(t = 2.00\ \text{ms} = 2.00 \times 10^{-3}\ \text{s}\), calculate:\[ F_{avg} = \frac{0.474\ \text{Ns}}{2.00 \times 10^{-3}\ \text{s}} \approx 237\ \text{N} \]

Key Concepts

Kinematic Equations

Kinematic equations are essential tools for solving problems related to the motion of objects, especially under constant acceleration. For a ball dropped from a certain height, these equations help us determine the velocity just before it strikes the ground. The relevant kinematic equation in this scenario is given by:\[ v_i = \sqrt{2gh_1} \]Here, \( v_i \) is the velocity just before impact, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( h_1 \) is the initial height from which the ball is dropped. By plugging in the known values, we calculate the velocity of the ball right before it impacts the surface.

• Initial height, \( h_1 = 2.00 \, \text{m} \)
• Gravity, \( g = 9.8 \,\text{m/s}^2 \)

Using these values, we compute:\[ v_i = \sqrt{2 \times 9.8 \times 2.00} \approx 6.26 \, \text{m/s} \]This calculation provides crucial information for further analysis, as it allows us to evaluate the change in velocity during the ball's impact with the slab.

Impact Force Calculation

The impact force during a collision can be determined using the concept of impulse. Impulse quantifies the change in momentum that occurs when an object is subjected to a force over a period of time. In the context of the steel ball, we calculate the impulse and use it to determine the average force exerted on the ball during its contact with the steel slab.The impulse \( J \) for the ball is calculated using the formula:\[ J = m(v_f + v_i) \]Where \( v_f \) is the velocity of the ball after rebounding, considered positively in the upward direction, and \( v_i \) is its velocity during impact.

• Mass of ball, \( m = 0.040 \, \text{kg} \)
• Initial velocity, \( v_i \approx 6.26 \, \text{m/s} \)
• Final velocity, \( v_f \approx 5.60 \, \text{m/s} \)

By applying these values, we find:\[ J = 0.040 \times (5.60 + 6.26) \approx 0.474 \, \text{Ns} \]The average force \( F_{avg} \) can then be determined using:\[ F_{avg} = \frac{J}{t} \]where \( t = 2.00 \, \text{ms} = 2.00 \times 10^{-3} \, \text{s} \). Substituting, we get:\[ F_{avg} = \frac{0.474}{2.00 \times 10^{-3}} \approx 237 \, \text{N} \]This result represents the magnitude of the average force exerted on the ball during impact with the slab.

Motion Under Gravity

When analyzing the motion of objects like our steel ball under gravity, it's crucial to understand how gravity affects their movement. Gravity is a force that attracts two bodies towards each other, and in this case, it's the force that pulls the ball downward.The analysis begins with the object dropping freely under the influence of gravity from a certain height. The initial velocity of such an object is zero, and it's accelerated by gravity alone.

• Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
• Initial velocity is zero when an object is free-falling.

As it falls, the velocity increases due to the gravitational pull until the point of impact when the ball first hits the slab. The principle of conservation of energy or kinematic equations can be used to track how potential energy (due to the height) converts into kinetic energy (associated with the motion). Post-impact, the ball still carries some kinetic energy, which propels it upward until it begins to succumb to gravity's pull again, resulting in its rebound height being lesser than the initial drop height.