Problem 11 - College Physics

Question

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a \(0.400 \mathrm{~kg}\) ball that is traveling horizontally at \(10.0 \mathrm{~m} / \mathrm{s}\). Your mass is \(70.0 \mathrm{~kg}\). (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so that afterward it is moving horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?

Step-by-Step Solution

Answer

(a) 0.0568 m/s; (b) 0.103 m/s.

Step 1: Understanding Conservation of Momentum

In problems involving collisions, especially when friction is negligible, we apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Step 2: Setting Up the Equation for Part (a)

Before the collision, the momentum of the system (you and the ball) can be calculated as only the ball has momentum because you are initially at rest. The momentum of the ball is given by \( m_b \cdot v_b \), where \( m_b = 0.400 \, \text{kg} \) and \( v_b = 10.0 \, \text{m/s} \). The momentum initially is therefore \( 0.400 \, \text{kg} \cdot 10.0 \, \text{m/s} = 4.0 \, \text{kg} \cdot \text{m/s} \).

Step 3: Applying Conservation of Momentum for Part (a)

After catching the ball, both you and the ball move together. Let your velocity after catching the ball be \( v \). The total mass of you and the ball is \( 70.0 \text{ kg} + 0.400 \text{ kg} = 70.4 \text{ kg} \). Using conservation of momentum, \( 4.0 \text{ kg} \cdot \text{m/s} = 70.4 \text{ kg} \cdot v \). Solving for \( v \), we get \( v = \frac{4.0}{70.4} \approx 0.0568 \, \text{m/s} \).

Step 4: Setting Up the Equation for Part (b)

For the situation where the ball bounces off with a velocity in the opposite direction, we again apply conservation of momentum. The ball's initial momentum remains \( 4.0 \, \text{kg} \cdot \text{m/s} \). Its final momentum, after bouncing back with \( 8.0 \, \text{m/s} \) in the opposite direction, is given by \( 0.400 \, \text{kg} \cdot (-8.0 \, \text{m/s}) = -3.2 \, \text{kg} \cdot \text{m/s} \).

Step 5: Applying Conservation of Momentum for Part (b)

Let \( v_Y \) be your velocity after the collision. According to conservation of momentum, the total momentum after the collision is: \( 70.0 \, \text{kg} \cdot v_Y - 3.2 \, \text{kg} \cdot \text{m/s} = 4.0 \, \text{kg} \cdot \text{m/s} \). Solving for \( v_Y \), we find:\[70.0 \, \text{kg} \cdot v_Y = 4.0 \, \text{kg} \cdot \text{m/s} + 3.2 \, \text{kg} \cdot \text{m/s}\]\( 70.0 \, \text{kg} \cdot v_Y = 7.2 \, \text{kg} \cdot \text{m/s} \)Thus, \( v_Y = \frac{7.2}{70.0} \approx 0.103 \, \text{m/s} \).

Key Concepts

Physics Problem Solving

Physics problems often require methodical solutions to ensure comprehension and accuracy. For problems involving collisions, such as this one, the key principle we utilize is the conservation of momentum. When friction is negligible, as is the case here, the momentum before a collision is equal to the momentum after the collision.
Solving these problems involves several critical steps:

• Identify the bodies involved and their initial momentum.
• Apply the conservation of momentum principle.
• Write down the equations representing the problem.
• Solve these equations to find the unknowns, distinct aspects like velocity after collision, etc.

Mastering these steps is essential for resolving complex situations in physics, making the problems manageable and solvable.

Collisions and Rebounds

Collisions are interactions where two or more bodies exert forces on each other for a significant duration, influencing each other's motion. In this problem, we consider two scenarios: catching the ball and letting it bounce off.
Understanding these types of interactions is critical:

• **Elastic Collision (Rebound):** The kinetic energy is conserved, and the colliding objects bounce off each other.
• **Inelastic Collision (Catching):** The objects stick together, and kinetic energy is not conserved, though momentum is conserved.

The exercise depicts both situations:
- In part (a), the process of catching the ball is an inelastic collision where you and the ball move as one post-collision.
- In part (b), the ball bouncing off represents an elastic collision where it rebounds with some speed while you move in the opposite direction.

Momentum Calculation

The key concept of momentum is essential in solving collision problems. It is defined as the product of an object's mass and its velocity. For effective momentum calculation:

• Determine the initial momentum of all objects. Initially, you have zero momentum since you start at rest, and the ball's momentum is given by its mass and velocity.
• Apply the rule of conservation: initial total momentum equals final total momentum.
• For part (a), calculate the system's combined mass and solve for the final velocity using the total initial momentum.
• For part (b), adjust for the momentum change as the ball rebounds and again solve for your resulting velocity.

These calculations help you understand how momentum is transferred during collisions.

Ice Friction Negligibility

Friction, the resistance to motion, is often a crucial factor in physics problems. However, in this scenario, the ice provides an ideal situation by offering negligible friction. This makes our calculations simpler.

• Friction's absence means the only forces at play are those involved in the collision itself.
• Thus, all of the momentum equations become straightforward without additional frictional forces to consider.
• As there is no loss of momentum to friction, anything that changes in the system is purely due to the collision, allowing for precise calculations and predictions.

Negligible friction offers a powerful platform to clearly observe physical principles like conservation of momentum in action, without the distraction of complex resistive forces.