1
$\begingroup$

trying to design an experiment for linear momentum. Could someone tell me whether this would work.

Get a wooden toy car and attach it to some string which is hung over a pulley on the end of the table. The other end of the string has a fixed mass at the end. As the mass is fixed the rate of change of momentum of the trolley is fixed. The length the trolley will travel over is 1m. Light gates will be used to measure the velocity of the trolley half way along (0.5m). I will change the mass of the car by placing masses on top of it. My question - if I measure the velocity at the light gate will the product of mass x velocity be the same if I keep the applied force the same? Or do I have to keep the time the same as well?

Thanks

$\endgroup$
1
$\begingroup$

You need to keep the amount of time the same as well. The change in momentum of a system is equal to impulse delivered to it, which is just the time-integral of the force: $$ \Delta \vec{p} = \vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \vec{F} \Delta t \text{ (if $\vec{F}$ is constant with respect to $t$.)} $$ So just having $\vec{F}$ constant isn't enough; you need to have $\Delta t$ constant as well.

Also: if you do the full analysis for the system you're describing, you'll find that the final velocity of the cart will not be $$ v_\text{trolley} = \frac{F \Delta t}{m_\text{trolley}} $$ but rather $$ v_\text{trolley} = \frac{F \Delta t}{ m_\text{trolley} + m_\text{hanger}}. $$ This is effectively because some of the impulse delivered to the system will go into making the hanger move along with the trolley. The easiest way to prove this rigorously is just to do the free-body diagrams for the trolley and the hanger, under the constraint that they both move with the same acceleration.

$\endgroup$
  • $\begingroup$ Thanks for the help, that makes sense. What experiment could I do using the equipment stated where I can investigate how mass affects momentum. I want to plot a graph of mass against some other variable in the equation. What could I do? $\endgroup$ – user37250 Jun 14 '15 at 14:13
  • $\begingroup$ When I teach intro physics labs, I have the students do a collision lab to illustrate conservation of momentum. I could envision an experiment where one cart comes in with a fixed amount of momentum (rolling down a ramp or launched by a spring or something), hits a second cart and locks together with it, and then you measure the final speed of the two carts. Since the incident momentum is constant, the final velocity should be inversely proportional to the combined mass of the carts. $\endgroup$ – Michael Seifert Jun 15 '15 at 15:33
1
$\begingroup$

I use this arrangement in my introductory classes when time allows, but not for momentum. I call it the "work-energy mini-lab". (It's a mini-lab because I don't expect a detailed write-up and guide the class through some of the harder analysis.)

Some things to note.

  • By measuring how far the mass hanger has to drop you get the distance over which the force is applied.
  • The force on the cart is not the weight dangling over the side but the tension in the string $$ T = \frac{m_c m_h}{m_c + m_h}g \,,$$ because the weight on the hanger has to serve to accelerate both masses.
  • That makes the work on any given run $W = T \cdot h$ where $h$ is the starting height of the hanger.
  • You should do your velocity measurement after the hanger has hit the ground, but not too far after or friction will cut into your precision.
  • I also vary the weight on the hanger rather than that of the cart.
  • I ask the students to plot $v^2$ against $T$, $h$, and $W$. You could ask them to do $v^2-vs-$m_h$ as well to show that this is not linear, but it is already getting pretty long for a mini-lab.

I also use the 5 meter air track we have rather than carts, but as long as your carts are good that is an implementation detail.

This really is a more natural for energy than momentum because you are able to measure the distance over which the force is applied directly, but that doesn't mean you can't get the momentum. It's just more computation and less direct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.