Elastic and un-elastic collisions

In summary, the bullet traveling at a speed of 300 m/s collides with a stationary carbon nucleus of mass 12m, and finds the velocity of the centre of mass to be 23.07 m/s. However, it is unclear whether you are supposed to be finding the velocity of the centre of mass before or after the collision. I will assume before.
  • #1
BananaMan
41
0
dear god i just typed a full page and it decided to give me a 404, i hate computers.

First time i had to use this forum in a long time but I am very stuck today and can't find a way out of it after over an hour and with reading my notes so here goes.

1)a proton of mass m traveling at 300m/s collides with a stationary carbon nucleus of mass 12m, the collision is elastic and head-on.

(a) find the velocity of the centre of mass of the system, for this i took an imaginary particle of mass 13m and used conservation of momentum to find a velocity of 23.07m/s ...not sure if that is right
(b) find the velocity of the proton after the collision in the centre of mass reference frame I've tried playing around with conservation of momentum for a while and with conservation of kinetic energy but my answer seems to always come out at either 0 or 300, not got a clue what I am doing wrong
(c)find the velocity of the proton after the collision in the lab reference frame i really am stuck on how to do this with conservation of momentum giving me one equation but 2 un-knowns


2)a bullet of mass m1 is fired with a speed v into the bob of a ballistic pendulum of mass m2. find the maximum height h attained by the bob if the bullet passes through the bob and emerges with a speed v/2 unfortunately i was hungover during this lecture and ended up falling asleep... whoooops, but it was a birthday night before so how could i not?

if anyone could point me in the right direction with these it would be greately appreciated

thanks
 
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  • #2
BananaMan said:
(b) find the velocity of the proton after the collision in the centre of mass reference frame I've tried playing around with conservation of momentum for a while and with conservation of kinetic energy but my answer seems to always come out at either 0 or 300, not got a clue what I am doing wrong
(c)find the velocity of the proton after the collision in the lab reference frame i really am stuck on how to do this with conservation of momentum giving me one equation but 2 un-knowns
been looking at these some more (at a more reasonable hour) but I am very confused, logically i would attempt to find the velocity of the proton in the lab reference frame possibly using m1/m2 = (v4-v2)/(v3-v1) however the order in which the questions are ordered makes me think that i am missing something very simple any advice pls?
 
  • #3
Are you given the equation for the velocity of centre of mass, or do you have to derive it yourself?
 
  • #4
i have typed everything i am given but as far as i am aware to get the velocity of the centre of mass all you do is have your "before" diagram as normal then on your "after" diagram you just draw a single imaginary particle with the complete mass of the system and work out the velocity of that, which is how i came to the conclusion of 23.07m/s
 
  • #5
(a) Your value is correct. However it is unclear whether you are supposed to be finding the velocity of the centre of mass before or after the collision. I will assume before.
(b) You can now use sum of velocities to determine the initial velocities of the proton and carbon atom in the centre of mass frame (i.e. subtract 23.7 m/s from each velocity). The determine the resultant velocity of the proton in that frame. (You have the relevant equation?)
(c) Then add the velocity of the centre of mass back to the photon's resultant velocity.
 
  • #6
El Hombre Invisible said:
(b) You can now use sum of velocities to determine the initial velocities of the proton and carbon atom in the centre of mass frame (i.e. subtract 23.7 m/s from each velocity). The determine the resultant velocity of the proton in that frame. (You have the relevant equation?)

first of all thanks for your answer however, if i do subtract 23.7 from each velocity i will get a velocity of -23.7 for the mass m2, which is wrong as it will be traveling in the same direction as the proton was originally

to calculate the resultant velocity of the proton in that frame, assuming that 23.7 is positive and the 276.3 i would get from subtracting 23.7 from 300 is infact negative and the proton travels back the other way, as it should? then would it not just be 23.7m/s x 2 ? as i believe this is the relative velocity of the centre of mass of the system to the proton? and if not, no i don't have the equation and just looked through the motion of centre of mass section in my textbook and cannot find anything that seems to fit this problem exept the equation to calculate the velocity of the centre of mass

thanks again
 
  • #7
BananaMan said:
first of all thanks for your answer however, if i do subtract 23.7 from each velocity i will get a velocity of -23.7 for the mass m2, which is wrong as it will be traveling in the same direction as the proton was originally

? As the distance between the two particles decreases, both particles will be moving towards their common centre of mass: i.e. in opposite directions. Please note I said to subtract Vcm from both velocities to find their initial velocities. Then you can determine the resultant velocities.



BananaMan said:
to calculate the resultant velocity of the proton in that frame, assuming that 23.7 is positive and the 276.3 i would get from subtracting 23.7 from 300 is infact negative and the proton travels back the other way, as it should? then would it not just be 23.7m/s x 2 ? as i believe this is the relative velocity of the centre of mass of the system to the proton?

If I follow you correctly, then yes you are right. If you have not been given this equation (and are not expected to derive it):

[tex]v_{1} = \frac{u_{1}(m_{1} - m_{2}) + 2m_{2}u_{2}}{m_{1} + m_{2}}[/tex]

then you can approximate by saying that the change in m2's velocity is negligible therefore:

[tex]v_{1} = -u_{1} + 2v_{2}[/tex]
 
  • #8
El Hombre Invisible said:
[tex]v_{1} = \frac{u_{1}(m_{1} - m_{2}) + 2m_{2}u_{2}}{m_{1} + m_{2}}[/tex]

now I am gettin somewhere, didnt really want to approximate so i used this equation to give me v1 = -253.84m/s in the CM reference frame

then adding back on my 23.07m/s (Vcm) i get the final velocity of the proton to be -276.9 in the lab reference frame

could you just check those over for me

thanks again for your help

sorry re - read and re-ran the equations realising i was using actual velocities as my inputs and found that v1=-276.916 in CM and then -300 in lab reference frame

which proves your approximation equation that it is roughly equal to the negative of the original velocity in the lab reference frame
 
Last edited:
  • #9
BananaMan said:
now I am gettin somewhere, didnt really want to approximate so i used this equation to give me v1 = -253.84m/s in the CM reference frame
then adding back on my 23.07m/s (Vcm) i get the final velocity of the proton to be -276.9 in the lab reference frame
could you just check those over for me
thanks again for your help
sorry re - read and re-ran the equations realising i was using actual velocities as my inputs and found that v1=-276.916 in CM and then -300 in lab reference frame
which proves your approximation equation that it is roughly equal to the negative of the original velocity in the lab reference frame

Show me your working. I think you've gone wrong somewhere. Having compared the propert equation with the approximation myself, I would say the approximation isn't justified here. (I got a difference of 50 m/s!)
 
  • #10
El Hombre Invisible said:
Show me your working. I think you've gone wrong somewhere. Having compared the propert equation with the approximation myself, I would say the approximation isn't justified here. (I got a difference of 50 m/s!)

actually re-running it i get a v1 in the CM reference of -276.93

and then adding on the Vcm i get 253.86 this what you get?

the error was because i forgot about the minus sign
 
  • #11
That looks more like it, though I get -277.5 (but I'm not using a decent calculator):

[tex]v_{1} = \frac{276.3(1 - 12) + 2 * 12 * -23.7}{1 + 12} = -277.5[/tex]?
 
  • #12
El Hombre Invisible said:
That looks more like it, though I get -277.5 (but I'm not using a decent calculator):
[tex]v_{1} = \frac{276.3(1 - 12) + 2 * 12 * -23.7}{1 + 12} = -277.5[/tex]?

yep that's what i got, thanks a lot :)

sorted the second question aswell, thanks alot
 

What is the difference between elastic and un-elastic collisions?

Elastic collisions are those in which kinetic energy is conserved, meaning the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In contrast, un-elastic collisions are those in which kinetic energy is not conserved, and some of the kinetic energy is converted into other forms, such as heat or sound.

What are some examples of elastic and un-elastic collisions?

An example of an elastic collision is when two billiard balls collide on a pool table. An example of an un-elastic collision is when a car crashes into a wall, as some of the kinetic energy is lost in the form of heat and sound.

How can you determine if a collision is elastic or un-elastic?

One way to determine if a collision is elastic or un-elastic is by calculating the total kinetic energy before and after the collision. If the values are the same, then the collision is elastic. If the values are different, then the collision is un-elastic.

What factors affect the elasticity of a collision?

The elasticity of a collision can be affected by factors such as the nature of the materials involved, the speed and direction of the objects before the collision, and the angle at which the objects collide.

Can a collision be partially elastic?

Yes, a collision can be partially elastic, meaning that some of the kinetic energy is conserved while some is lost. This can occur in cases where the objects involved are slightly deformed or if there is friction present during the collision.

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