VERY HARD M2 question

OP

SsEe

You have two momentum equations, two restitution equations.

Unparseable latex formula:

4u = 4u_1 + 5v_1 \\[br]-4u_2 = 4u_1 - 5ev_1 \\[br]\frac{v_1 - u_1}{u} = e \\[br]\frac{u_2}{u_1+ev_1} = e [br]

I'll let you work out what everything stands for. Then just eliminate everything until you have an equation in just u and e. u is a common factor throughout which divides out and you're left with the required cubic.

easier said than done :s-smilie:

Reply 4

OP

nota bene

Isn't B) 'just' to solve the cubic equation and chose the root that is 0<e>=1 ?

A seems painful though...

err maybe, i didnt even look at b) because a) was so hard

Reply 5

westhamfan

easier said than done :s-smilie:

This isn't actually that difficult, you just need to be careful and methodical.

Reply 6

OP

lol ITS SO FRUSTRATING the variables just keep reappearing. If somebody could provide a full solution to part a) I would be very grateful. I worked out the equations myself but its just so annoying when you can't do it.

PLEASE HELP ME!!!

Reply 7

Solution as requested by PM. Excuse the little ASCII art diagrams.

Before first collision: --v--> A B
After first collision: <--a1--A B--b1-->

Cons momentum (1): 5b1-4a1 = 4u.
Law of restitution (2): a1+b1 = eu
(1)+4 x (2): 9b1=(4e+4)u
(2) x 5 - (1): 9a1=(5e-4)u

Before collision with wall: <--a1--A B--b1-->
After collision with wall: <--a2--A B<--b2--

b2 = eb1, so 9b2 = (4e^2+4e)u. Nothing happened to a, so a2 = a1.

Before final collision: <--a2--A B<--b2--
After final collison: <--a3--A
Cons momentum: 4a3 = (4a2+5b2)u = (20e-16+20e^2+20e)u/9 = 4(5e^2+10e-4)u/9
Law of restitution (4): a3 = e (b2-a2) = e(4e^2+4e-5e+4)u/9 = (4e^3-e^2+4e)u/9

Equating the two expressions for a3 gives:

So (4e^3-e^2+4e) - (5e^2+10e-4) = 0. That is, 4e^3-6e^2-6e+4 = 0, or 2e^3-3e^2-3e+2 = 0.

Reply 8

OP

DFranklin

Solution as requested by PM. Excuse the little ASCII art diagrams.

Before first collision: --v--> A B
After first collision: <--a1--A B--b1-->

Cons momentum (1): 5b1-4a1 = 4u.
Law of restitution (2): a1+b1 = eu
(1)+4 x (2): 9b1=(4e+4)u
(2) x 5 - (1): 9a1=(5e-4)u

Before collision with wall: <--a1--A B--b1-->
After collision with wall: <--a2--A B<--b2--

b2 = eb1, so 9b2 = (4e^2+4e)u. Nothing happened to a, so a2 = a1.

Before final collision: <--a2--A B<--b2--
After final collison: <--a3--A
Cons momentum: 4a3 = (4a2+5b2)u = (20e-16+20e^2+20e)u/9 = 4(5e^2+10e-4)u/9
Law of restitution (4): a3 = e (b2-a2) = e(4e^2+4e-5e+4)u/9 = (4e^3-e^2+4e)u/9

Equating the two expressions for a3 gives:

So (4e^3-e^2+4e) - (5e^2+10e-4) = 0. That is, 4e^3-6e^2-6e+4 = 0, or 2e^3-3e^2-3e+2 = 0.

absolute legend.

Reply 9

OP

DFranklin

Cons momentum: 4a3 = (4a2+5b2)u

where has the u come from at the end of the equation?

Reply 10

westhamfan

where has the u come from at the end of the equation?

Sorry, it's a typo. It shouldn't be there.

Reply 11

OP

thanks mate thats great

Reply 12

Rawlo

The solution relies on the fact that after the first collision ball A changes its direction of motion. Is this definetely the case??

RoRaw

Reply 13

Rawlo

The solution relies on the fact that after the first collision ball A changes its direction of motion.Actually, it doesn't. Nothing to stop one of the velocities being negative if needed.

Reply 14

Mohit_C

10

Lol, I made another thread on this question. It's a bit of a pain really.

Reply 15

Swayum

If you thought that question was hard, try the one after it. A whole school's maths department couldn't do it so they e-mailed our school for help. It took my mechanics teacher like 4 sides of work or something to get the solution. Funny thing is, while he was explaining how to do THIS question in class, he couldn't do it even though he managed it by himself during the holidays.

Reply 16