# Newton's Third Law Problem

1. Mar 31, 2005

### hya_been

So I have a homework question that says:
A 100 kg hockey player a 112kg hockey player collide with each other each travelling with a force of 50N.

The text book answer says that the acceleration of the 112kg is 0.89m/s^2
and the acceleration of the 100kg player is -1.0 m/s^2.

I'm confused because according to Newton's third Law there will be equal and opposite reaction forces so wouldn't they not move? Is the text book wrong or am I?

Last edited: Mar 31, 2005
2. Mar 31, 2005

### whozum

Remember by the second law:

$$F = m a$$

They both impart a force of 50N on each other, but the acceleration is also scaled by their masses. 50N on a 100kg person will push them farther than 50N on a 112kg person.

Imagine pushing a shopping cart with 50N. An empty shopping cart will go really far, whereas a full shopping cart will go a much shorter distance. In both instances the force is the same, but:

$$a = \frac{F}{m}$$

3. Mar 31, 2005

### hya_been

calculating acceleration

So you do 100/100 and 100/112 even though the total of the forces equals 0 because its 50 + -50??

4. Mar 31, 2005

### whozum

The net force of the entire system is 0, but the force on each player is of magnitude 100N. You can find the acceleration of each player using

$$a = \frac{F}{m}$$. Knowing F = 100N and m = 100kg for person 1,

$$a = \frac{100N}{100kg} = 1.0m/s^2$$

F = 100N and m = 112kg for person 2:

$$a = \frac{100N}{112kg} = 0.89m/s^2$$

5. Mar 31, 2005

### hya_been

Thank You!!

Thank-you you are a saviour!