# Thread: Collisions and Many Particle Systems

1. ## Collisions and Many Particle Systems

Is anybody able to help out with the attached questions please?

2. How much have you managed to complete?

Tell me where you are stuck and I can help you. I'd feel uncomfortable just giving a set of answers to your homework.

3. I can't do any of part (a). My mind just goes blank when I look at this question.

Moreover, this is not homework. I'm trying to study mechanics by myself and this is a question I found without solutions.

4. OK, for a (i) can you see how the lengths are related, right?

From the diagram you should be able to see that

$\boxed{y+l = d + x}$

by equating the distances that get you from the wall to the mass $m_1$ .

ii. &iii. What are the salient features here:

1. The string over the pulley transmits the same tension force, $T$ to each mass.
2. The fact that $m_1$ and $m_2$ are connected in this way means they must have the same acceleration.
3. The table is smooth so there's no friction forces on $m_2$ .

So for the mass $m_1$ you must have:

$\boxed{m_1 a = m_1 g - T}$ [1]

where $g$ is the acceleration due to gravity and $a$ is the inertial acceleration. This is from Newton II.

Using Newton II for mass $m_2$ we must have:

$\boxed{m_2 a = T - F_s}$ [2]

where $F_s$ is the force due to the extension of the spring.

I hope the diagrams should be obvious from the above. As for the pulley the forces you have are the tension $T$ acting away from the pulley at each corner and reaction force diagonally outwards from the pulley (at a 45 degree angle to the horizontal).

The only force equation left is that of the spring where:

$\boxed{F_s = k (y- l_0 )}$ [3].

Sub [3] into [2] and you get:

$\boxed{m_2 a = T - k( y - l_0 )}$ [4]

and now add [1] to [4] to give you

$( m_1 + m_2 ) a = m_1 g - k ( y - l_0 )$,

rearranging this and substituting for y gives us:

$\boxed{a = \frac{m_1 g - k (d+x -l - l_0 )}{m_1 + m_2}}$ .

Hope that was of some help!