Question

Physics help? I know it's the wrong area but there are always more active people here.


I have two tables, each with ten trials of elastic and inelastic collisions. Each table has the two objects mass and velocities before and after the collision. I'm supposed to calculate the momentum for each trial of the objects individual momentum before and after the trials, and then their momentum together for the inelastic trials. I hope that makes sense so far.

My concern comes in when I'm calculating the momentum for all of the inelastic collisions. Each object has its own momentum before the collision, but the momentum of the combined objects is literally just the addition of their momentum separately. For example, object 1 has a momentum of .4Ns and object 2 has a momentum of .15Ns. The momentum after the collision for both objects combined is .55Ns.

I feel like this is incorrect because they should both lose energy when they collide, but I'm not sure if I should be factoring that into these calculations at all.

I just want to make sure that this is right before I type it all into a data table and finish it. So if anyone can confirm/deny/help that'd be great.

Thanks.

Answer 1

If the energy is relatively small like 0.15N the momentum lost will be insignificant. Thats what I think :)

Answer 2

.4Ns
.15Ns
.55Ns

I feel like I can help, but first
what are these units exactly? is N for newtons? or is N a vvariable for mass or velocity?

Answer 3

Ns, or kg m/s is the unit of momentum.

Answer 4

kgxm/s, actually

Answer 5

alright, just checking ^_^ thanks

Answer 6

No problem, So, do you think these are right then? My teacher mentioned that we'll learn more about factoring in the amount of energy lost when we're in more advanced areas of physics, but I don't know if these are right even without that.

Answer 7

and inelastic collisions mean the two objects, basically combine right?
so we would right it as:
\[m_1(v_1) + m_2(v_2) = (m_1+m_2)v_{final}\]

do you have the values of the objects mass and initial velocities or do you just have the .4, .15, .55 values?

Answer 8

i have everything.

m1=.8kg
v1(initial) =.5m/s
m2 = .5kg
v2(initial)=.3
v(final)=.423

Answer 9

Left side of the equation comes out to .55, right side comes out to 1.01

Answer 10

Nope, I'm wrong. Typed it into my calculator wrong. Didn't use parenthesis. It's even.

Answer 11

so, we'd have:
.8(.5) + .5(.3) = E + (.8+.5).423
.55 = E + .5499
where we will let E represent the amount of energy that is lost due to friction and vibration of the actual collision

when we solve for E, we get
E = 0.0001

so in this collision, a very small amount of energy is lost. which is reasonable.

Answer 12

Yep, Thank you though! I didn't remember that formula as a way to check all of the work. Thank you very much.

Answer 13

^_^

to answer your question, i'd say you've done it right

Answer 14

Thank you! Have a great night.