91Ó°ÊÓ

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?

Step by step solution

01

Calculate the final velocities of basketball and golf ball

As both balls fall freely under gravity, we use the following kinematic equation to find their final velocities (assuming down is positive): \(v^2 = u^2 + 2as\) where \(u\) is the initial velocity (\(0\) in this case since they start from rest), \(v\) is the final velocity, \(a\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)) and \(s\) is the distance fallen (we consider only the basketball's height, \(0.701 \mathrm{m}\)). \(v^2 = 0^2 + 2(9.81)(0.701)\) \(v = \sqrt{13.76}\) \(v \approx 3.71 \mathrm{m/s}\) The final velocity of basketball and golf ball is approximately \(3.71 \mathrm{m/s}\) downwards, just before hitting the ground.

02

Compute absolute values of momenta before collision

The momenta (absolute values) of both balls just before hitting the ground can be calculated using their masses and their final velocities: \(momentum_{ball} = m_{ball} \times v_{ball}\) \(momentum_{golf} = m_{golf} \times v_{golf}\) For the basketball, we compute the momentum by: \(momentum_{ball} = (0.619)(3.71)\) \(momentum_{ball} = 2.29 \mathrm{kg\cdot m/s}\) For the golf ball, we compute the momentum by: \(momentum_{golf} = (0.0459)(3.71)\) \(momentum_{golf} = 0.1704 \mathrm{kg\cdot m/s}\) The absolute values of the momenta are \(2.29 \mathrm{kg \cdot m/s}\) and \(0.1704 \mathrm{kg \cdot m/s}\) for the basketball and golf ball, respectively.

03

Apply conservation of momentum and coefficient of restitution for the collisions

In a completely elastic collision, the coefficient of restitution, \(e\), is \(1\). Therefore, the approach velocity is equal to the separation velocity. Mathematically, \(e = \frac{v_{2,after} - v_{1,after}}{v_{1,before} - v_{2,before}}\) Let \(v_{1,after}\) be the final velocity of the basketball after the collision, and \(v_{2,after}\) be the final velocity of the golf ball after the collision. Since the floor is considered fixed during the collision, the final velocity of the basketball after the collision with the floor is the opposite of its initial velocity, \(v_{1,before} = -3.71 \mathrm{m/s}\). We can now apply conservation of momentum, knowing that the golf ball's initial momentum was \(0.1704 \mathrm{kg \cdot m/s}\).

04

Calculate absolute magnitude of the momentum of the golf ball after the collisions

Applying conservation of momentum for the golf ball and the basketball, we have: \(momentum_{golf,before} + momentum_{ball,before} = m_{golf}v_{2,after} - m_{ball}v_{1,after}\) Substituting the values, we get: \(0.1704 + 2.29 = (0.0459)(v_{2,after}) - (0.619)(-3.71)\) Rearrange the equation and solve for \(v_{2,after}\), the final velocity of the golf ball after the collision: \(v_{2,after} = 21.2 \mathrm{m/s}\) (positive sign, upward direction) The absolute magnitude of the momentum of the golf ball after the collisions is \(momentum_{golf,after} = m_{golf} \times v_{2,after}\) \(momentum_{golf,after} = (0.0459)(21.2)\) \(momentum_{golf,after} = 0.9727 \mathrm{kg \cdot m/s}\)

05

Determine the height the golf ball reaches after the collision

We can now use its final velocity of \(21.2 \mathrm{m/s}\) to calculate the maximum height it reaches above the ground. As it goes up, it decelerates due to gravity until it stops momentarily before falling. Using the following equation: \(v^2 = u^2 - 2as\) where \(v\) is the final velocity (\(0\ \mathrm{m/s}\) when stopped), \(u\) is the initial velocity (after the collision), \(a\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), and \(s\) is the distance covered upward. Plugging in the values: \(0^2 = (21.2)^2 - 2(9.81)s\) \(s = \frac{(21.2)^2}{2(9.81)}\) Calculate the distance yielded: \(s \approx 22.7\ \mathrm{m}\) Note that the golf ball went up 22.7 meters from the collision point with the basketball, but it was already 0.701 meters above the ground at the point of the collision. The total height from the ground is therefore: \(height = 22.7 + 0.701 = 23.401\ \mathrm{m}\) Thus, the golf ball will bounce up to approximately \(23.4\ \mathrm{m}\) above the ground.

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Key Concepts

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

Momentum Conservation

One of the most fundamental concepts in physics is the conservation of momentum. In an isolated system, the total momentum before an event, such as a collision, is the same as the total momentum after the event—momentum is conserved. This principle is crucial when analyzing collisions, including the engaging example of a golf ball and basketball collision from our textbook exercise.

Momentum, a vector quantity, is described by the product of an object's mass (\(m\)) and its velocity (\(v\)). In the case of the basketball and golf ball, each falling separately before impact, their individual momenta can be calculated using their respective masses and velocities just before collision. But, it's when they interact—collide—that momentum conservation really plays a role. For a totally elastic collision, which the problem assumes, both momentum and kinetic energy are conserved. This situation allows us to predict post-collision velocities once we understand the pre-collision momenta of the objects involved.

In the step-by-step solution, we computed the momentum of the basketball and the golf ball just before the collision with the ground and calculated the change in momentum for the golf ball after impacting the basketball. The fascinating outcome—astronomically high rebound height for the golf ball—is a direct consequence of momentum conservation applied during the collisions.

Kinematic Equations

Kinematic equations are indispensable tools for describing the motion of objects in physics. They provide a mathematical framework to relate an object’s displacement, acceleration, velocity, and the time of travel. With these equations, one is able to predict future motion based on initial conditions, assuming acceleration is constant.

The problem at hand involves a type of motion known as free fall, where the only acceleration acting on the objects is that due to gravity \(a = 9.81 \text{m/s}^2\). The kinematic equation \(v^2 = u^2 + 2as\) allows us to calculate the final velocity (\(v\)) of both the basketball and the golf ball as they reach the ground, given that they start from rest (\(u = 0\)) and fall from a known height (\(s\)).

Understanding and correctly applying these kinematic formulas are key in solving the exercise. After the golf ball collides with the basketball, we use another kinematic equation to determine how high the golf ball will rise post-collision. Here, we predict the stopping point (maximum height) of the golf ball using its upward velocity immediately after the collision and the constant acceleration (deceleration in this case) due to gravity.

Energy Conservation in Collisions

In physics, the conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. During elastic collisions, like the one between the basketball and golf ball, kinetic energy—the energy of motion—is particularly considered. In an ideally elastic collision, the total kinetic energy of the system before collision is equal to the total kinetic energy after collision.

This exercise illustrates energy conservation by using the initial heights and masses of the balls to determine their velocities just before impact—their kinetic energies at that moment are directly related to these velocities. After the collision, because the collision is elastic and energy is conserved, we can calculate the final velocities of the objects involved. Specifically, the surprising height achieved by the golf ball, determined in step 5 of the solution, directly results from the kinetic energy obtained during the collision with the basketball.

It is important to note that while all elastic collisions conserve momentum, not all momentum-conserving collisions are elastic. Inelastic collisions, in contrast to elastic ones, do not conserve kinetic energy even though momentum is still conserved. For the purpose of our exercise, however, we’ve assumed a perfectly elastic collision, where both momentum and kinetic energy conservation principles apply.