# A-level Physics conservation of momentum question help

Watch
Announcements
#1
The question is divided in two parts, and I managed to work out part ii) as it was fairly straight forward. The question is:

A particle A, of mass m, moving with an initial velocity u makes a head on collision with particle B of mass 2m, being initially at rest. In terms of u calculate the final velocity of A if the collision is i) elastic ii) inelastic. Assume that in the inelastic collision the two particles adhere.

I am not quite sure of how to go on to solve part i) as we don't have any info regarding the final KEs or partcle B's final velocity. Am I thinking too much into it and missing something very obvious?

PS: I have tried to solve it by simultaneous equations by calling x and y, A and B's respective final velocities into the equations for conservation of energy and momentum but ended up with the wrong answers.

0
5 years ago
#2
In an elastic collision kinetic energy is conserved, in an inelastic collision it isn't.
0
5 years ago
#3
(Original post by Joules911)
The question is divided in two parts, and I managed to work out part ii) as it was fairly straight forward. The question is:

A particle A, of mass m, moving with an initial velocity u makes a head on collision with particle B of mass 2m, being initially at rest. In terms of u calculate the final velocity of A if the collision is i) elastic ii) inelastic. Assume that in the inelastic collision the two particles adhere.

I am not quite sure of how to go on to solve part i) as we don't have any info regarding the final KEs or partcle B's final velocity. Am I thinking too much into it and missing something very obvious?

PS: I have tried to solve it by simultaneous equations by calling x and y, A and B's respective final velocities into the equations for conservation of energy and momentum but ended up with the wrong answers.

In part ii) it says that the two particles adhere which means that the resultant mass becomes the sum of the individual masses. Hence the total mass becomes 3m.

Momentum before collision = momentum after collision

So mu = 3mv, so v=u/3

For part i) the particles do not coalesce. Remember that for elastic collisions, the speed at which two bodies approach one another is the same as the speed at which the two bodies move away from one another. Hence, uB - uA = vA - vB

Since uB = 0, the equation becomes -uA = vA - vB

Therefore, uA = vB - vA

Now, we have to use the conservation of momentum:

muA = mvA + 2mvB

uA = vA + 2vB

vB = (uA - vA)/2

Simply plug VB into equation 1 to find vA:

uA = (uA - vA)/2 - vA

2uA = uA - vA - 2vA

3va = -uA

Hence va = -uA/3

This means that when A collides with B, it rebounds (since it has opposite direction due to -ve sign) and the speed of A after the collision is 1/3 of its original speed.

Hopefully this helps.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### What support do you need with your UCAS application?

I need help researching unis (24)
13.48%
I need help researching courses (12)
6.74%
I need help with filling out the application form (9)
5.06%
I need help with my personal statement (75)
42.13%
I need help with understanding how to make my application stand out (41)
23.03%
I need help with something else (let us know in the thread!) (3)
1.69%
I'm feeling confident about my application and don't need any help at the moment (14)
7.87%