- #1

Remi David

- 2

- 1

**Hello !**

## Homework Statement

Two blocks M

_{1}and M

_{2}are connected by a string of negligible mass. If the system is released from rest, find how far block M

_{1}slide in time ##t##. Neglect friction.

__Diagram__:

See Attached Image

__Clue given in the manual__:

If M

_{1}= M

_{2}, then solution is ##x(t)= \frac{gt^2}{4}##

## Homework Equations

[/B]

__Newton's Laws__:

##\sum_{}^{} \vec{F} = m\vec{a}##

## \vec{F}_{m1/m2} = -\vec{F}_{m2/m1} ##

__Relation between ##x(t)##, ##\vec{v}(t)##, ##\vec{a}(t)##__:

##\vec{v}(t) = \frac{dx}{dt}##

##\vec{a}(t) = \frac{dv}{dt}##

## The Attempt at a Solution

So I started the exercise by separating the two masses M

_{1 }and M

_{2}, and I did a diagram with all the forces exerting on each masses.

Then I used my two Newton's laws, to find out the relation between:

1) The tension generated on each masses (Newton third law)

2) Acceleration of M

_{2}based on the tension generated by M

_{2}

__Here is my approach__:

I wrote down:

##\sum_{}^{} \vec{F}_{M2} = m_{M2}\vec{a}_{M2}##

##\sum_{}^{} \vec{F}_{M1} = m_{M1}\vec{a}_{M1}##

So, with ##\vec{T}## being the tension applied to the string, and ##\vec{R}## the resistance from the table:

##\vec{T}_{M2} - \vec{W}_{M2} = m_{M2}\vec{a}_{M2}##

##\vec{R}_{M1} - \vec{W}_{M1} + \vec{T}_{M1} = m_{M1}\vec{a}_{M1}##

I simplified the equation for M

_{1}, by suppressing ##\vec{R}_{M1}## and ##\vec{W}_{M1}## as ##\vec{R}_{M1} - \vec{W}_{M1} = 0##

I concluded that tension generated by M

_{2}was the tension applied on M

_{1}

So,

##\vec{T}_{M2} = - \vec{T}_{M1}##

##\vec{T}_{M1} = - m_{M2}\vec{g}##

using that relation, we can write:

## \vec{a}_{M1} = \frac{m_{M2}}{m{_M1}}g##

Differentiating , I found that:

##\vec{v}(t) = v_{0} + \frac{m_{M2}}{m{_M1}}gt ## with ## v_{0} = 0##

##\vec{x}(t) = x_{0} + \frac{m_{M2}}{2m{_M1}}gt^2 ## with ## x_{0} = 0##

As per the clue given in the manual, I should have a "4" instead of a "2" on the last equation... Looks like my mistake is coming for my differentiation, but I have no idea why... The "4" will come when differentiating a ##x^3##.

Don't give me the answer, I just want to have clues as I'm looking to improve (I'm not involved in any kind of scholarship at this time, and I'm doing that for my personal pleasure :D).

Sorry if my English is not the best, English is not my first language (but I'm trying to improve as well)

Thanks

Have a good day

Rémi