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Chapter 7: Problem 90

A soccer ball with mass \(0.265 \mathrm{~kg}\) is initially at rest and is kicked at an angle of \(20.8^{\circ}\) with respect to the horizontal. The soccer ball travels a horizontal distance of \(52.8 \mathrm{~m}\) after it is kicked. What is the impulse received by the soccer ball during the kick? Assume there is no air resistance.

Short Answer

Expert verified
The soccer ball receives an impulse of approximately 6.75 kg鈰卪/s during the kick.

Step by step solution

01

Determine the initial velocity

To find the impulse, we first need to know the initial velocity of the soccer ball. Since there is no air resistance, we are dealing with a projectile motion problem. The horizontal distance traveled by the soccer ball can be described by the following equation: $$x = v_0 \cos\theta \cdot t$$ Where \(x\) is the horizontal distance, \(v_0\) is the initial velocity, \(\theta\) is the angle at which the ball was kicked, and \(t\) is the time of flight. We can also find an expression for the time of flight using the vertical motion equation: $$y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2$$ Since the soccer ball lands on the ground, we have \(y=0\). By solving this equation for \(t\) and substituting it back into the horizontal distance equation, we can get an expression for the initial velocity.
02

Calculate the time of flight

First, let's solve the vertical motion equation for \(t\): $$0 = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2$$ The quadratic formula gives us: $$t = \frac{2v_0 \sin\theta}{g}$$
03

Calculate the initial velocity

Now, we substitute the expression for \(t\) from Step 2 into the horizontal distance equation: $$x = v_0 \cos\theta \cdot \frac{2v_0 \sin\theta}{g}$$ $$52.8 \mathrm{~m} = v_0^2 \sin20.8^{\circ} \cos20.8^{\circ} \cdot \frac{2}{9.81 \mathrm{~m/s^2}}$$ Solving for the initial velocity, we get: $$v_0 = \sqrt{\frac{52.8 \mathrm{~m} \cdot 9.81 \mathrm{~m/s^2}}{2 \sin20.8^{\circ} \cos20.8^{\circ}}} \approx 25.46 \mathrm{~m/s}$$
04

Calculate the impulse

Now that we have the initial velocity, we can calculate the impulse, which is given by the change in momentum: $$I = m \Delta v = m(v_f - v_i)$$ The soccer ball was initially at rest, so \(v_i = 0 \mathrm{~m/s}\). The final velocity is the initial velocity after the kick, which is \(v_f = 25.46 \mathrm{~m/s}\). The mass of the soccer ball is \(0.265 \mathrm{~kg}\). Therefore, the impulse is: $$I = (0.265 \mathrm{~kg}) (25.46 \mathrm{~m/s} - 0 \mathrm{~m/s}) \approx 6.75 \mathrm{~kg \cdot m/s}$$ So, the impulse received by the soccer ball during the kick is approximately \(6.75 \mathrm{~kg \cdot m/s}\).

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Key Concepts

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

Projectile Motion
Projectile motion refers to the two-dimensional motion of an object that is launched into the air and subject to only gravitational acceleration. A key characteristic of projectile motion is that the horizontal and vertical motions are independent of each other, except for the time they take. This means that the horizontal velocity remains constant since there are no additional forces in the horizontal direction (assuming no air resistance), while the vertical velocity changes because of gravity.

When an object is launched at an angle, the initial velocity vector (\(v_0\)) has both horizontal and vertical components. These components can be found using trigonometry: the horizontal component is \(v_0 \cos(\theta)\), and the vertical component is \(v_0 \sin(\theta)\), where \(\theta\) is the launch angle. The object will follow a parabolic trajectory and eventually return to the same vertical level it was launched from, which in the case of the soccer ball is when it touches the ground again (\(y = 0\)).
Initial Velocity Calculation
Determining the initial velocity is pivotal for solving projectile motion problems. The initial velocity can be calculated if you know the distance traveled, the angle of launch, and the acceleration due to gravity (\(g\)). The previous explanation provides a method for determining the initial velocity of a soccer ball in such a scenario by using the horizontal range of the projectile and the time of flight.

In situations where air resistance can be neglected, calculating the initial velocity involves first finding the time of flight using the vertical motion equation. You can use the derived time of flight to substitute into the horizontal distance formula. This process eventually allows you to solve for the initial velocity. The calculation reveals the speed at which the projectile must move to cover the desired distance, considering the influence of gravity on its vertical motion.
Momentum Change
Momentum, in physics, is the quantity of motion of a moving body and is calculated as the product of mass and velocity (\(p = mv\)). The change in momentum, or the impulse, occurs whenever a force is applied over a time interval. Impulse has the same units as momentum (\(\mathrm{kg \cdot m/s}\)) and is given by the change in momentum of an object.

The formula for impulse (\(I\)) is essentially the mass of the object multiplied by the change in velocity (\(\Delta v\)), which is \(I = m(v_f - v_i)\). It's significant to remember that momentum is a vector quantity, so direction is essential: an impulse can alter either the magnitude or the direction of the momentum, or both. In the scenario of the soccer ball, the ball initially at rest receives an impulse that corresponds to the ball's subsequent motion, determined by the final velocity right after the kick.

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Most popular questions from this chapter

When hit in the face, a boxer will "ride the punch"; that is, if he anticipates the punch, he will allow his neck muscles to go slack. His head then moves back easily from the blow. From a momentum-impulse standpoint, explain why this is much better than stiffening his neck muscles and bracing himself against the punch.

The value of the momentum for a system is the same at a later time as at an earlier time if there are no a) collisions between particles within the system. b) inelastic collisions between particles within the system. c) changes of momentum of individual particles within the system. d) internal forces acting between particles within the system. e) external forces acting on particles of the system.

A sled initially at rest has a mass of \(52.0 \mathrm{~kg}\), including all of its contents. A block with a mass of \(13.5 \mathrm{~kg}\) is ejected to the left at a speed of \(13.6 \mathrm{~m} / \mathrm{s} .\) What is the speed of the sled and the remaining contents?

Here is a popular lecture demonstration that you can perform at home. Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground. (For reasons that should become clear once you solve this problem, do not attempt to do this experiment inside, but outdoors instead!) With a little practice, you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). a) If the balls are released from a height where the bottom of the basketball is at \(0.701 \mathrm{~m}\) above the ground, what is the absolute value of the basketball's momentum just before it hits the ground? b) What is the absolute value of the momentum of the golf ball at this instant? c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions? d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball?

A 60.0 -kg astronaut inside a 7.00 -m-long space capsule of mass \(500 . \mathrm{kg}\) is floating weightlessly on one end of the capsule. He kicks off the wall at a velocity of \(3.50 \mathrm{~m} / \mathrm{s}\) toward the other end of the capsule. How long does it take the astronaut to reach the far wall?
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