Two masses on an inclined plane without friction

[RiotRick]

Homework Statement

Two identical masses are connected with a rope and are gliding without any friction. Situation given in the picture:

Determine after which distance "s" they stop if we have s=0 at t=0 with starting velocity $v_0$

Given:
$\alpha$ and $\beta$ with $\alpha < \beta$
$v_0$, $m$,$g$

Homework Equations

No friction

The Attempt at a Solution

I guess they mean when the two masses stop for the very first time, since both masses are equal and $\beta$ is bigger than alpha, they will start gliding towards the right side.
Say the left mass is m1 and the right mass m2 and say $v_0$ is positive.
Then:
(1)$g*m_1*sin(\alpha)-F_{rope}=m1*a$
(2)$-g*m_2*sin(\beta)+F_{rope}=m2*a$
Solving 2nd equation for F and using the fact bot masses are equal, we get in the first equation:
$g*m*sin(\alpha)-m*a-g*m*sin(\beta)=m*a$
$a=\frac{g}{2}*(sin(\alpha)-sin(\beta))$

Now how do I continue from here? I tried to use integrate twice $s = \frac{gt^2}{4}*(sin(\alpha)-sin(\beta))+v_0*t+0$ ($s_0=0$) but then how the get rid of the time?

 

[scottdave]

Are you familiar with the equations of motion for constant acceleration? Specifically, the one V = V₀ + a*t should help you. Find the time at which V = 0.

 

[PeroK]

Homework Statement

Two identical masses are connected with a rope and are gliding without any friction. Situation given in the picture:
View attachment 236305
Determine after which distance "s" they stop if we have s=0 at t=0 with starting velocity $v_0$

Given:
$\alpha$ and $\beta$ with $\alpha < \beta$
$v_0$, $m$,$g$

Homework Equations

No friction

The Attempt at a Solution

I guess they mean when the two masses stop for the very first time, since both masses are equal and $\beta$ is bigger than alpha, they will start gliding towards the right side.
Say the left mass is m1 and the right mass m2 and say $v_0$ is positive.
Then:
(1)$g*m_1*sin(\alpha)-F_{rope}=m1*a$
(2)$-g*m_2*sin(\beta)+F_{rope}=m2*a$
Solving 2nd equation for F and using the fact bot masses are equal, we get in the first equation:
$g*m*sin(\alpha)-m*a-g*m*sin(\beta)=m*a$
$a=\frac{g}{2}*(sin(\alpha)-sin(\beta))$

Now how do I continue from here? I tried to use integrate twice $s = \frac{gt^2}{4}*(sin(\alpha)-sin(\beta))+v_0*t+0$ ($s_0=0$) but then how the get rid of the time?

You have calculated a constant acceleration. Can you solve the problem without introducing an unknown time variable?

 

[RiotRick]

You have calculated a constant acceleration. Can you solve the problem without introducing an unknown time variable?

I don't know how.

 

[PeroK]

I don't know how.

There is another equation you could use. But, if you don't know it you could follow the advice of @scottdave to calculate and eliminate $t$ from your equation.

You seem to me to have done the hard part of the problem already.

 

[RiotRick]

Are you familiar with the equations of motion for constant acceleration? Specifically, the one V = V₀ + a*t should help you. Find the time at which V = 0.

I get $s=\frac{v_0^2}{2a}+\frac{v_0^2}{a}$
Replacing my a gives me:
$s=\frac{1}{2}*\frac{v_0}{\frac{g*(sin(\alpha)-sin(\beta)}{2}}+\frac{2*v_0}{g*(sin(\alpha)-sin(\beta)}$
$s=\frac{3*v_0}{g*(sin(\alpha)-sin(\beta)}$
which is wrong by the factor 3.

 

[PeroK]

I get $s=\frac{v_0^2}{2a}+\frac{v_0^2}{a}$
.

I'm not sure how you got that equation.

 

[gneill]

Hint: There's another SUVAT equation that involves initial and final velocities, acceleration, and distance. No time involved.