Question

1. a car moving to the right applies 500 n of force to a bug that hits the windshield. what is the magnitude and direction of the reaction force from the bug back onto the windshield? 2) if a soccer ball with a mass of 0.45 kg is kicked with a force of 60 n, what will its acceleration be? 3) how much force would a model rocket with a mass of 0.25 kg need to reach an acceleration of 75 m/s²?

Explanation:

1)

Step1: Apply Newton's third law

Newton's third law states that for every action, there is an equal and opposite reaction.
The force exerted by the car on the bug (\(F_{car - bug}=500\space N\)) and the force exerted by the bug on the car (\(F_{bug - car}\)) are an action - reaction pair.

Step2: Determine magnitude and direction

Since \(F_{bug - car}=-F_{car - bug}\) (the negative sign indicates direction), the magnitude of the force is \(500\space N\) and the direction is to the left.

2)

Step1: Use Newton's second law \(F = ma\)

We need to solve for \(a\), so we can rewrite the formula as \(a=\frac{F}{m}\)
Given \(F = 60\space N\) and \(m=0.45\space kg\)

Step2: Calculate acceleration

Substitute the values into the formula: \(a=\frac{60}{0.45}=\frac{60}{\frac{9}{20}}=\frac{60\times20}{9}=\frac{1200}{9}\approx133.33\space m/s^{2}\)

3)

Step1: Use Newton's second law \(F = ma\)

Given \(m = 0.25\space kg\) and \(a = 75\space m/s^{2}\)

Step2: Calculate force

Substitute the values into the formula: \(F=(0.25)(75)= 18.75\space N\)

Answer:

1. Magnitude: \(500\space N\), Direction: to the left
2. \(a\approx133.33\space m/s^{2}\)
3. \(F = 18.75\space N\)