# Angular Momentum vs Linear Momentum

1. Dec 4, 2014

### Maged Saeed

1. The problem statement, all variables and given/known data
The following figure shows an overhead view of a thin rod of mass M=2.0 kg and length L = 2.0 m which can rotate horizontally about a vertical axis through the end A. A particle of mass m = 2.0 kg travelling horizontally with a velocity $$v_i=10 j \space m/s$$ strikes the rod (which was initially at rest) at point B. the particle rebounds with a velocity $$v_f=-6 j\space m/s$$. Find the angular speed of the rod just after the collision.

2. Relevant equations

$$(I\omega)_i=(i\omega)_f$$

3. The attempt at a solution

I have tried to solve this question using the previous equation , but I'm stuck with the momentum of the ball . Should it be linear or angular?

I mean ;
which of the following equation should i use :
1)
$$(mvl)_{ball}=(I\omega)_{rod} +mvl$$
$$(2 \times 2 \times 10j)=(\frac{2 \times 2^2}{12}+2 \times 1^2)+2\times 2 \times -6j$$
The moment of inertia of the rod is:
$$\frac{ml^2}{12}+mh^2$$
where h is the distance from the center of mass to the axis of rotation.

2)
$$(mv)_{ball}=(I\omega)_{rod}+mv$$
$$(2 \times 2)=(\frac{2 \times 2^2}{12}+2 \times 1^2)+(2 \times -6j)$$

The equation 2 seems to lead to the correct answer , but , why i should take the linear momentum instead of angular momentum!!

2. Dec 4, 2014

### BvU

Which equation has the same dimension for all terms ?

3. Dec 4, 2014

### Maged Saeed

Yeah , ..
How it comes that I didn't pay attention to this point ..

Thanks ..