Question

1. which object has greater momentum, a butterfly traveling at 16 km/h (10 mi/h) or an eagle traveling at 16 km/h (10 mi/h)?
2. describe how to recognize a transfer of momentum.
3. using what you know of momentum, describe the transfer of momentum that would happen if a car hit a skateboarder.

Explanation:

1.

Step1: Recall the momentum formula

Momentum \(p = mv\), where \(m\) is mass and \(v\) is velocity. Given \(v_{butterfly}=v_{eagle}\), and \(m_{eagle}>m_{butterfly}\) (since an eagle has a larger mass than a butterfly).

Step2: Compare momenta

Since \(p = mv\) and \(v\) is the same, \(p\propto m\). So \(p_{eagle}>p_{butterfly}\)

2.

A transfer of momentum occurs when there is a change in the motion of objects. You can recognize it by observing a change in velocity (speed and/or direction) of objects involved in an interaction. For example, when two billiard balls collide, the motion (velocity) of each ball changes, indicating a transfer of momentum.

3.

When a car hits a skateboarder, the car has a large momentum (\(p = mv\), with large \(m\) and some \(v\)). The skateboarder has much less momentum initially. During the collision, the car transfers some of its momentum to the skateboarder. The skateboarder’s velocity (and thus momentum) will increase (they will be pushed forward), and the car’s velocity (and thus momentum) will decrease (it will slow down a bit), as momentum is transferred from the car to the skateboarder.

Answer:

1. The eagle has greater momentum.
2. Observe changes in the velocity (speed and/or direction) of objects in an interaction.
3. The car transfers some of its momentum to the skateboarder. The skateboarder’s velocity (and momentum) increases (they are pushed forward), and the car’s velocity (and momentum) decreases (it slows down a bit).