Question

1. a 30 kg girl and a 40 kg boy stand on rollerskates, face one another, and push off each other. if the boy accelerates backward with an acceleration of 6 m/s², what will be the acceleration of the girl?
2. a 1000 kg car collides with a 1 mg mosquito. by what factor will the acceleration of the mosquito be greater than the acceleration of the car when they collide?

Explanation:

Step1: Use Newton's third law

According to Newton's third law, the force exerted by the girl on the boy \(F_{gb}\) is equal in magnitude and opposite in direction to the force exerted by the boy on the girl \(F_{bg}\), i.e., \(F_{gb}=-F_{bg}\). The magnitude \(|F_{gb}| = |F_{bg}|\).

Step2: Apply Newton's second law \(F = ma\)

For the boy, \(F_{gb}=m_{b}a_{b}\), where \(m_{b} = 40\space kg\) and \(a_{b}=6\space m/s^{2}\). So \(F_{gb}=40\times6 = 240\space N\).
For the girl, \(F_{bg}=m_{g}a_{g}\), and since \(|F_{bg}|=|F_{gb}|\), we have \(a_{g}=\frac{F_{bg}}{m_{g}}\). Substituting \(F_{bg} = 240\space N\) and \(m_{g}=30\space kg\), we get \(a_{g}=\frac{240}{30}\).

Answer:

\(8\space m/s^{2}\)