Problem 72 - College Physics

Question

A 60-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 25 m/s. (a) If the player was moving upward with a speed of 4.0 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? (b) If the ball is in contact with the player’s head for 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

Step-by-Step Solution

Answer

The velocity of the ball after impact will be approximately 51.1 m/s upwards and the average acceleration of the ball is approximately 3805 m/s^2.

Step 1: Applying the principles of an elastic collision

An elastic collision is one in which kinetic energy is conserved. The equation \(m1*v1 + m2*v2 = m1*u1 + m2*u2\) can be used to describe elastic collisions, where m1 and m2 are the masses of the two objects, v1 and v2 are their respective velocities before collision, and u1 and u2 are their velocities after collision. Let's denote the soccer player as object 1 and the soccer ball as object 2. Initially, the soccer player is jumping upwards with a speed of 4.0 m/s, thus v1 = 4.0 m/s. The ball is coming downwards with speed 25 m/s, so v2 = -25 m/s (negative as it's downwards). The soccer player has a mass of 60 kg and the ball 0.45 kg. So we're missing u2, the final velocity of the soccer ball.

Step 2: Solving for final velocity of the soccer ball

Let's rearrange the equation \(m1*v1 + m2*v2 = m1*u1 + m2*u2\) to solve for u2. We know that after the collision, the player is still moving upwards, let's say with speed u1, and the ball rebounds, moving upward with speed u2. We're neglecting the force of gravity as the duration of the collision is very small, so u1 = v1 = 4.0 m/s. Substituting known values into the equation and solving for u2, we get \( (60 kg * 4.0 m/s) + (0.45 kg * -25 m/s) = (60 kg * 4.0 m/s) + (0.45 kg * u2)\), thus simplifying we find that u2 = 51.1 m/s.

Step 3: Calculating average acceleration of the ball

For part (b), we're asked to compute the average acceleration of the ball which is the rate of change of velocity. This can be calculated using the equation \(a = Δv/Δt\), where Δv is the change in velocity and Δt is the time for which the ball was in contact with the player, in this case 20 ms or 0.02 s. The change in velocity is the final velocity minus the initial velocity. Given that the ball was moving downward initially, we take this as negative, so Δv = u2 - (-v2) = u2 + v2 = 51.1 m/s + 25 m/s = 76.1 m/s. Using this, we find that \(a = 76.1/0.02 = 3805 m/s^2\).

Key Concepts

Conservation of Kinetic Energy

Understanding the principle of conservation of kinetic energy is crucial when analyzing elastic collisions in physics. This fundamental concept asserts that in an ideal elastic collision, the total kinetic energy before and after the event remains unchanged. That is, no kinetic energy is lost to other forms of energy, like heat or sound, which is usually the case in real-world scenarios.

To delve deeper, during an elastic collision, the mathematical expression to encapsulate the conservation of kinetic energy is \( \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 \), where \(m_1\) and \(m_2\) represent the masses of the colliding bodies, while \(v_1\), \(v_2\), \(u_1\), and \(u_2\) symbolize their velocities before and after the collision, respectively. However, in homework problems, such as the one with the soccer player and ball, some external forces like gravity are negligible during the brief moment of impact, simplifying our calculation to a great extent.

When students tackle problems involving elastic collisions, it's essential to not only apply this principle but also thoroughly understand it. By recognizing that kinetic energy is conserved, they can set up equations that enable them to solve for unknown quantities, like the final velocities of the objects involved in the collision.

Average Acceleration

The concept of average acceleration comes into play when we're looking at the change in an object's velocity over time. In physics, acceleration describes how quickly an object speeds up, slows down, or changes direction. It's represented by the equation \( a = \frac{\Delta v}{\Delta t} \) where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval over which this change occurs.

Applying this to our example, we see how the soccer ball experiences a dramatic increase in speed from a downward to an upward trajectory within a very short time frame, which results in a very high average acceleration. These large accelerations are typical of collisions where contact is made for just a moment, known in sports as an 'impact'. It's important to comprehend that the term 'average' denotes a constant acceleration over a specified time period, which may not reflect the actual variation of acceleration during the collision. Calculating average acceleration provides us with an overall understanding of the forces involved without needing a complex analysis of the interaction's mechanics.

For students, grasping the measure of average acceleration deepens their understanding of how forces act during collisions and the immediate after-effects on the objects' motion, especially in situations where detailed force models are not available or necessary.

Momentum Conservation

Momentum conservation is another foundational principle in collision physics. Momentum, the product of an object's mass and velocity, is a vector quantity that provides an immense insight into the motion of objects. The law of conservation of momentum states that, for a closed system of objects that interact with each other, the total momentum before the interaction equals the total momentum after, provided no external forces are acting on the system.

In the context of the soccer player and ball, when they collide, the total momentum before contact must be the same as after they separate. The momentum conservation equation \( m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 \) lets us calculate missing variables, such as the ball's final speed after the player heads it. Key to solving these problems is setting a consistent direction as positive — typically upwards — and then ensuring all velocities are directed accordingly, with the opposite direction being negative.

This concept is especially important for students to master because it reveals the inherent 'balance' in systems where no external forces are involved, allowing predictions about the post-collision velocities of objects. In-depth knowledge of momentum conservation empowers students to tackle a broad range of physical problems beyond the textbook, fostering a more profound comprehension of dynamic interactions.