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A ball of mass \(0.500 \mathrm{~kg}\) is dropped from a height of \(2.00 \mathrm{~m}\). It bounces against the ground and rises to a height of \(1.40 \mathrm{~m} .\) If the ball was in contact with the ground for \(0.0800 \mathrm{~s}\), what average force did the ground exert on the ball?

Short Answer

Expert verified
The average force exerted by the ground on the ball is obtained by dividing the change in momentum by the time the ball was in contact with the ground. Substituting the values to the impulse equation provides the solution.

Step by step solution

01

Calculate initial velocity before hitting the ground

The initial velocity of the ball before hitting can be calculated by the formula \(v_i^2=2*g*h\), where \(g = 9.80 m/s^2\), the acceleration due to gravity and \(h = 2.00 m\), the height from which the ball is dropped. Substituting these values we get \(v_i = \sqrt{2*9.8*2.00}\) m/s.
02

Calculate final velocity after bouncing

The final velocity of the ball after bouncing can be calculated by the same formula \(v_f= \sqrt{2*g*h'}\), where \(g = 9.80 m/s^2\), the acceleration due to gravity and \(h' = 1.40 m\), the height to which the ball bounces. Substituting these values we get \(v_f = \sqrt{2*9.8*1.40}\) m/s.
03

Calculate the change in momentum

The change in momentum can be calculated as \(\Delta p = m*v_f - (-m*v_i)\) where \(m = 0.500 kg\), the mass of the ball, \(v_f\) is the final velocity and \(v_i\) is initial velocity as calculated in the previous steps. Substituting the values will give us the change in momentum.
04

Calculate the average force

The average force can be calculated by using the formula of impulse as \(F = \Delta p / \Delta t\), where \(\Delta p\) is the change in momentum as calculated in the previous step and \(\Delta t = 0.0800 s\), is the time the ball was in contact with the ground. Substituting these values will give us the average force.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impulse-Momentum Theorem
The impulse-momentum theorem is a fundamental concept in physics that relates an object's momentum change to the impulse applied to it. Impulse, which is the product of force and the time period over which it acts, equates to the change in momentum of an object. It's mathematically expressed as \( I = \triangle p \), where \( I \) represents impulse, and \( \triangle p \) represents the change in momentum.
In the exercise, the ball experiences a change in momentum during the short interval when it contacts the ground. Given that impulse is also equal to average force multiplied by contact time (\( F_{avg} \times \triangle t \)), we can rearrange the equation to solve for the average force as \( F_{avg} = \triangle p / \triangle t \).
Understanding this relationship allows us to see how the length of time the ball is in contact with the ground (the time interval) directly affects the average force exerted by the ground. A shorter contact time would lead to a larger average force, and vice versa.
Conservation of Momentum
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on it. In the context of collisions and interactions like the ball hitting the ground, the total momentum just before impact should equal the total momentum just after impact if we ignore external influences like air resistance.
When applying this concept to our exercise, we must notice that the ground exerts a force on the ball, changing its momentum from the initial value before the bounce to the final value after the bounce. However, the ground itself reacts with an equal and opposite force according to Newton's third law, ensuring the overall system—if including the Earth—still conserves momentum.
Kinetic Energy and Work
Kinetic energy is the energy of motion. An object with mass (\( m \)) moving at velocity (\( v \)) has kinetic energy expressed as \( KE = \frac{1}{2}mv^2 \). Work, in physics, refers to the process of energy transfer to or from an object via the application of force along a displacement.
In our example, as the ball falls, it converts gravitational potential energy into kinetic energy, which is then partly transferred to the ground and back to the ball as it bounces. When the velocity of the ball changes as a result of the force applied by the ground, work is done on the ball. This concept doesn't directly calculate the force but gives us insight into the energy transformations during the bounce.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. The GPE of an object is calculated by the product of its weight (mass times gravitational acceleration) and its height above a reference point, as given by \( GPE = mgh \).
The initial GPE of the ball, when held at a height of 2.00 meters, is what propels it downwards upon release. As it falls, the GPE is converted into kinetic energy. After the bounce, it doesn't return to its original height because some energy is lost, likely as sound and heat, and possibly due to air resistance. The height to which it bounces back gives us a means of calculating its speed just after the bounce, which in turn is used to determine the change in momentum and subsequently the average force.

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Most popular questions from this chapter

A pitcher claims he can throw a \(0.145\) -kg baseball with as much momentum as a \(3.00-\mathrm{g}\) bullet moving with a speed of \(1.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) What must the baseball's speed be if the pitcher's claim is valid? (b) Which has greater kinetic energy, the ball or the bullet?

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A cue ball traveling at \(4.00 \mathrm{~m} / \mathrm{s}\) makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of \(30.0^{\circ}\) with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

A ball of mass \(0.150 \mathrm{~kg}\) is dropped from rest from a height of \(1.25 \mathrm{~m}\). It rebounds from the floor to reach a height of \(0.960 \mathrm{~m}\). What impulse was given to the ball by the floor?

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter (d) the Earth (mass running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\)

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