Search More Than 6,000,000 Posts

# Thread: Ok physics(collision calculations) buffs take a look

1. ## Ok physics(collision calculations) buffs take a look

I have this problem. In the before(före)pic the box m is about to collide with the left 2m box. After the first collision the m box bounces back and collides with the 2m box to the right. My task is to calculate the speed of m+2m aftre they move togheter

the first collision is completely elastic and the second is completely inelastic.

The thing is thats just 2 easy calculations and I get the result that the end speed of m+2m is v/6(v is the begining speed of the m block before any collision). m offocurse represents mass.

Our proffesor most often gives us insanely tricky task and this one seems to easy. Am I missing something?

first I did this calculation

Vx=v*(m-2m)/(m+2m)=v*(-m)/(3m)=-v/3
Vx is speed of the m block after first collision

then I did this to calculate speed of m+2m after second collision.
I changed the direction of the x-axis btw to get a positive value of Vx

I used this
mAvA + mBvB=(mA+mB)V2

mA=m vA=Vx mB=2m and vB=0

so I have V2= (mVx)/(m+2m) and because Vx=v/3 I get =>

V2= (v/3)*(1/3)=v/6

So the answere is v/6. Does this look right?

2. LOL..its hard to understand how you wrote it, I'm assuming this is an impulse problem, and you used the right equation

m1v1 (before collison) = m2v2 (after collison)

(mass 1st object times velocity 1st object) = (mass 2nd object times velocity of 2nd object) Energy is always conserved.. pE=kE

Manipulate the formula to solve for the unknown..

It is a tricky question.. You can look at the two boxes (m,2m on the right) as one object, they are traveling at the same velocity after the collison because the way he explained the problem..

Use the velocity of box m, before the colision in the before part of the equation, and calculate the velocity after the colision of (2m,m)

Is that the exact picture of the problem you attached..

whats all the given information:

mass of box 2m on the left:
*velocity of this is zero, i'm assuming
mass of box m on the right:
velocity of box m on the right, in the before picture:
mass of box 2m on the right in the before picture:

and

what was the given information in the picture after the collison?

mass/velocity of box 2m on the left
mass/velocity of boxes m2m on the right

I couldn't understand the info you gave, i'll work the problem out, and compare what i got with you..

3. all boxes except the middle one is standing still. after first collision the left 2m box gets speed to the left and the middle one bounces and heads for the box 2m to the right that is not moving until they colide.

the m symbolises the mass. and v is the speed. no numerical values at all. So box 2m has mass 2m and box m has mass m

that is the exact picture of the problem minus the text he sent with it(im guessing you dont know swedish anyway lol). but the only important things in the text is that the first collision is completely elastic and the second one is completely inelastic.

thanks for the help bro!

4. Well... It's simple if i understood you correctly.. The first colision has nothing to do with the problem because its elastic.. disregard it bounced off the first 2m box on the left.. Use the momentum formula to calculate the velocity (speed) of box m2m on the right..

There is no given information on the diagram?

5. Originally Posted by monster.
Well... It's simple if i understood you correctly.. The first colision has nothing to do with the problem because its elastic.. disregard it bounced off the first 2m box on the left.. Use the momentum formula to calculate the velocity (speed) of box m2m on the right..

There is no given information on the diagram?
No info on the diagram, he wants a non numerical answere. Just a expression for the velocity based on the begining velocity.

there is no given info on mass or speed anywhere. The V3m you have asked about is the speed I am suposed to calculate.

Box m has initial velocity V and mass m, both boxes 2m has initial velocity 0 and mass 2m.

Isnt the first collision important for the problem? Even though its a elastic collision the energy or momentum doesnt stay the same for m? I mean 2m gains speed and momentum from the collision so m must lose some? Cause I need to give the answere based on the initial velocity then the first collision cant be disregarded

Because I know the momentum(mv) of m before it collides then I have to calculate how much momentum m gets in the oposit direction after the first collision.

Im using the momentum conservation formula for elastic collision that you included with the file for the first collision and then I use this inelastic colision formula for the second collision (m1v1+m2v2=(m1+m2)*v cause then those 2 bodys act as one body with the combined mass m+2m.
If v1 is the speed of m to the right after the first collison and cause v2=0 I and I know m1=m and m2=2m I can simplify it to look like this m1v1=(m+2m)v
Where v is the final speed of the 2 blocks togheter.
Rewriten that becomes v=(mv1)/(m+2m)
I can rewrite that to
v=(mv1)/3m. I can then eliminate m and I have left is (1/3)*v1.
If v1 is (1/3)v (v beeing original speed) like I can get out of the calculation for the elastic collision I then have
(1/3)*(1/3)v=(1/9)*v. That should be the final answere(I mistakenly wrote v/6 in my first post)???
In other words V3m=v/9

Hope it makes more sense now Im not trying to find a numerical value like say 6m/s. I just need to find out how big the combined speed of m+2m(to the right) is after both those collisions compered to the original speed of m.
Last edited by Kärnfysikern; 02-12-2004 at 02:52 PM.

6. Ok i'm understanding you a lil better bro.. Physics is all the same, no matter where you from.. I hate to keep questioning now, but you said m is accelerating after the first colision? You will need to incorporate the initial velocity (from the first colision) to calculate the velocity of m2m.. 3 step problem, same formula.. but the final velocity of part 1 'before' will be used as the initial velocity in the 'after' scenerio.. The final answer looks good, let me throw a dip in, and get back to this thread in a lil bit bro..

I just got off work!!!!

post whore time!!!!!!!!!!!!!!!!!!!!!!!!!!!11111

7. u lost me after u said i have this problem....

8. Originally Posted by monster.
Ok i'm understanding you a lil better bro.. Physics is all the same, no matter where you from.. I hate to keep questioning now, but you said m is accelerating after the first colision? You will need to incorporate the initial velocity (from the first colision) to calculate the velocity of m2m.. 3 step problem, same formula.. but the final velocity of part 1 'before' will be used as the initial velocity in the 'after' scenerio.. The final answer looks good, let me throw a dip in, and get back to this thread in a lil bit bro..

I just got off work!!!!

post whore time!!!!!!!!!!!!!!!!!!!!!!!!!!!11111
Yeah thats right!
well m is not accelerating after the first question(or well it obviously accelerates upp to the speed that can be caluclated from the momentun equations but that is negligable cause we can find out the velocity without thinking about acceleration)

Newtons laws and mechanics must be the most boring parts of physics blah I cant wait to dig into relativity and quantum physics!