Calculating Initial Velocity in Elastic Collision of Two Balls on a String

[STEMucator]

Homework Statement

A $0.02kg$ ball is fired horizontally with a speed $v_0$ towards a $0.1kg$ ball hanging motionless from a $1.0m$ long string.

The balls undergo a head-on, perfectly elastic collision, after which the $0.1kg$ ball swings out to a maximum angle $\theta = 50°$.

What was $v_0$?

Homework Equations

$m_1 = 0.02kg$
$m_2 = 0.1kg$
$L = 1.0m$
$\theta = 50°$
$v_0 = ?$

The Attempt at a Solution

So I figured right at the beginning of the system (the instant the little ball collides with the big one), there would be kinetic + gravitational energy. By the time the bigger ball has reached it's maximum height, all of the energy is in the form of gravitational energy.

$E_i = E_f$
$E_{k_i} + E_{g_i} = E_{g_f}$
$\frac{1}{2}mv_i^2 + mgy_i = mgy_f$
$mv_i^2 = 2mg(y_f - y_i)$

The masses $m$ on both sides of this equation are not equal and so cannot be canceled I think. The mass on the left is the mass of the small ball I believe. The mass on the right is the mass of the larger ball. So I get:

$v_i = \sqrt{\frac{2m_2g(y_f - y_i)}{m_1}}$

To fill this equation in, conservation of momentum needs to be considered. Since the collision is perfectly elastic and the bigger mass is initially at rest:

$p_{T_i} = p_{T_f}$
$m_1v_{1_i} = m_1v_{1_f} + m_2v_{2_f}$
$v_{1_i} = \frac{m_1v_{1_f} + m_2v_{2_f}}{m_1}$

To finish the conservation of momentum, I believe I need these two equations:

$v_{1_f} = \frac{m_1 - m_2}{m_1 + m_2}v_0$
$v_{2_f} = \frac{2m_1}{m_1 + m_2}v_0$

So presuming I sub everything in and get a value for $v_{1_i}$, which will definitely include a $v_0$ term, I can plug that into the left side of my energy equation for $v_i$.

Now I may be mistaken, but I believe I need to use forces to determine the last piece of information I need, namely $y_f - y_i$. To find $y_i$, I should consider the bigger ball when it is initially at rest. Then:

$F_{net_y} = F_T - F_G \Rightarrow F_T = F_G = m_2g$

To find $y_f$ I should consider the bigger ball at its maximum height. I believe I'll need to use components for this.

Before I continue though, I'm hoping this is on the right track. Thank you for any help in advance.

 

[voko]

What happens with the small ball after collision? Does it lose all of its velocity?

 

[STEMucator]

No, it will not lose all of its velocity. It will reverse directions with a slight loss of speed, but will still be moving.

 

[voko]

So, not all of its energy is transferred to the big ball. I do not think this is taken into account in your equations.

I would further suggest that you label masses and velocities (to which ball they belong) in your equations from the beginning, otherwise it is very confusing.

 

[STEMucator]

So, not all of its energy is transferred to the big ball. I do not think this is taken into account in your equations.

I would further suggest that you label masses and velocities (to which ball they belong) in your equations from the beginning, otherwise it is very confusing.

Indeed, I was just making sure I had the right ideas before I started subbing things in.

So even though the small ball is not part of the "system" per say, it is still moving after the collision. So do I need to account for the final kinetic energy of the little ball anyway?

 

[voko]

Yes, the small ball must be taken into account.

 

[STEMucator]

Yes, the small ball must be taken into account.

So before I sub anything in, my energy equation would change to:

$E_i = E_f$
$E_{k_i} + E_{g_i} = E_{k_f} + E_{g_f}$
$\frac{1}{2}mv_i^2 + mgy_i = \frac{1}{2}mv_f^2 + mgy_f$
$0 = 2mg(y_f - y_i) + m(v_f^2 - v_i^2)$

The above equation is completely generic. Now I'll sub the actual masses in just to avoid confusion:

$0 = 2m_2g(y_f - y_i) + m_1(v_f^2 - v_i^2)$

Now the conservation of momentum portion of my first post would let me deal with the right hand term on the right side, namely:

$v_i = v_{1_i} = \frac{m_1v_{1_f} + m_2v_{2_f}}{m_1}$

$v_f = v_{1_f} = \frac{m_1 - m_2}{m_1 + m_2} v_0$

If this looks more reasonable, I'll concentrate on the forces.

 

[voko]

So before I sub anything in, my energy equation would change to:

$E_i = E_f$
$E_{k_i} + E_{g_i} = E_{k_f} + E_{g_f}$
$\frac{1}{2}mv_i^2 + mgy_i = \frac{1}{2}mv_f^2 + mgy_f$

After the collision, both balls have non-zero velocities. This is not reflected in the equation.

 

[STEMucator]

After the collision, both balls have non-zero velocities. This is not reflected in the equation.

Say what? The larger ball will have a velocity of zero at its maximum height won't it?

If it helps, I know the answer is listed as 7.9m/s.

 

[haruspex]

$v_f = v_{1_f} = \frac{m_1 - m_2}{m_1 + m_2} v_0$
.

That equation looks wrong to me. How did you derive it?
To get the right value for $v_{1_f}$ you either need to apply conservation of energy just to the collision or use Newton's experimental law.

 

[STEMucator]

That equation looks wrong to me. How did you derive it?
To get the right value for $v_{1_f}$ you either need to apply conservation of energy just to the collision or use Newton's experimental law.

For a system where one object is at rest and a perfectly elastic collision occurs, that equation can be used to determine the final speed of object 1, namely the one in motion.

Then in the energy equation (in the 'final' phase) the only 'velocity' would be the velocity of the ball that deflected. So there is only kinetic energy from the small ball that's still moving, while the larger ball is at its maximum height with a velocity of zero.

EDIT: So is the energy equation I wrote wrong? I'm very confused now with the feedback.

 

[haruspex]

For a system where one object is at rest and a perfectly elastic collision occurs, that equation can be used to determine the final speed of object 1, namely the one in motion.

Yes, sorry, my mistake.
And everything else looks fine to me. I think voko did not realize you were skipping consideration of the KE immediately after collision and going straight to the final energy.

 

[voko]

Indeed, because some key equations lacked labels denoting what ball a symbol belonged to, I got confused. Specifically, in #7, $$ {1 \over 2} mv_i^2 + m g y_i = {1 \over 2} mv_f^2 + m g y_f $$ had the same symbol $m$ everywhere, so it was not clear which ball it was. Now it seems it was used for both balls, which is a very bad thing to do. $v_i$ and $v_f$ were marginally better, because the message did say in the end that those were of the small ball, but I did not read that part of the message because the equations stopped making sense before.

Such confusion in symbols is not acceptable. If that were a homework assignment and I were evaluating it, I would reject it even if the final answer were correct. All the symbols used must be clearly explained, and no symbol should be used for more than one thing (not without a very good reason and an adequate description and warning).

 

[STEMucator]

Sorry about the confusion guys, I'm just used to working generically until I need to plug things in. I was trying to clarify which mass was which by subbing them in this step (post #7):

$0 = 2m_2g(y_f - y_i) + m_1(v_f^2 - v_i^2)$

This highlights the idea that only the gravitational energy of the big ball is being considered before the collision $(y_i)$ and in the instant it reaches it maximum height $(y_f)$.

Only the kinetic energy of the little ball is being considered as well. In the instant of the collision $(v_i)$ is the speed of the little ball. After the big ball has reached its maximum height, the speed of the little ball should be $(v_f)$.

I'm having a bit of trouble producing the answer of 7.9m/s though.

I figured $y_f = L(1 - cos \theta)$, but I'm not sure.

 

[voko]

Sorry about the confusion guys, I'm just used to working generically until I need to plug things in.

You seem to misunderstand what "generically" means. When, for example, you have two masses, and they both enter one equation, having one symbol for both is not generic. It is just plain wrong. When you have two masses, and each is governed by an equation that is independent of the other mass, and both equations are identical in the form, you could use one symbol for both masses, but you should then explain that very carefully.

I figured $y_f = L(1 - cos \theta)$, but I'm not sure.

That is correct.

 

[STEMucator]

You seem to misunderstand what "generically" means. When, for example, you have two masses, and they both enter one equation, having one symbol for both is not generic. It is just plain wrong. When you have two masses, and each is governed by an equation that is independent of the other mass, and both equations are identical in the form, you could use one symbol for both masses, but you should then explain that very carefully.
That is correct.

Here's a list of variables I'm plugging into the energy equation. These don't seem to produce the final answer:

$y_f = L(1-cos(50°))$
$y_i = L$ - Not sure about this one

$v_f = (0.67v_0)$
$v_i = (0.98v_0)$

 

[voko]

I am confused again. I thought $v_i$ was the velocity of the little ball just prior to the collision, and $v_f$ was its velocity after the collision. Then you should have $v_i = v_0$.

 

[STEMucator]

I am confused again. I thought $v_i$ was the velocity of the little ball just prior to the collision, and $v_f$ was its velocity after the collision. Then you should have $v_i = v_0$.

Yes sorry, I was looking at the conservation of momentum I did earlier, but I now realize it doesn't make sense to plug the value I found from it into the equation. I'm referring to:

$v_i = v_{1_i} = \frac{m_1v_{1_f} + m_2v_{2_f}}{m_1}$

So really $v_i = v_0$.

I'm pretty sure the final speed is correct though:

$v_f = v_{1_f} = \frac{m_1 - m_2}{m_1 + m_2} v_0$

Still no 7.9m/s though.

 

[voko]

Wait. The formula for $y_f$ is OK only if downward is positive, but I think the expression for the potential energy assumes upward is positive. Fix that. What is the final expression you have for $v_0$, symbolically?

 

[STEMucator]

Wait. The formula for $y_f$ is OK only if downward is positive, but I think the expression for the potential energy assumes upward is positive. Fix that. What is the final expression you have for $v_0$, symbolically?

The book I'm using uses the convention that +x and +y are positive. Re arranging the energy equation from prior:

$v_f^2 + \frac{2m_2g(y_f - y_i)}{m_1} = v_i^2$

$\sqrt{v_f^2 + \frac{2m_2g(y_f - y_i)}{m_1}} = v_0$

So I should also multiply my expression for $y_f$ by $(-1)$ to switch its direction.

 

[voko]

I agree with your equations so far. But I would like to see the formula with $y_f$ and $y_i$ already substituted.

 

[STEMucator]

I agree with your equations so far. But I would like to see the formula with $y_f$ and $y_i$ already substituted.

I'm a bit confused about how to interpret $y_i$ in this problem. Also, I thought only magnitudes and not the directions mattered for energy conservation?

Here's what I do know though:

$\sqrt{(0.67v_0)^2 + \frac{2*0.1*9.8(y_f - y_i)}{0.02}} = v_0$

 

[voko]

$y_i$ must be the initial height of the little ball. It depends on what you regard "zero" height (so does $y_f$, too).

Magnitudes enter kinetic energy. Potential energy depends on directions as well.

 

[STEMucator]

$y_i$ must be the initial height of the little ball. It depends on what you regard "zero" height (so does $y_f$, too).

Magnitudes enter kinetic energy. Potential energy depends on directions as well.

Hmm so what you're saying is I can make this a little easier by defining $y_i = 0$?

 

[voko]

I am not sure it will be easier, but you certainly can.

 

[STEMucator]

I am not sure it will be easier, but you certainly can.

Success!

By defining the initial position of the big/small ball to be the 'zero' of the gravitational energy, I get the right answer.$v_0 = \sqrt{v_f^2 + \frac{2m_2g(y_f - y_i)}{m_1}}$
$v_0 = \sqrt{ (0.66v_0)^2 + \frac{2*0.1*9.8 (1-cos(50°)-0) }{0.02} } = 7.87$

Only 2 significant figures allowed, therefore $v_0 = 7.9m/s$.

I believe the trick here was to define the zero of the gravitational energy and measure the increase in gravitational energy from the zero. Perhaps I should employ this trick more often as to reduce the workload.