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Coefficient of restitution

Two smooth balls A and B of equal radius and mass 4kg and 5kg respectively lie at rest on a smooth horizontal floor. Ball Ais projected with speed u and collides with ball B. Following this collision ball B then strikes a smooth vertical wall normally. After rebounding from the wall, ball B again collides with ball A. Given that ball B is brought to rest by this second collision with A,

a) show that 2e^3 - 3e^2 - 3e + 2 = 0

where e is the coefficient of restitution throughout the collisions.

b) verify that e=1/2 is the only practical solution of the equation.

If you solve this you are immense at maths lol

Two smooth balls A and B of equal radius and mass 4kg and 5kg respectively lie at rest on a smooth horizontal floor. Ball Ais projected with speed u and collides with ball B. Following this collision ball B then strikes a smooth vertical wall normally. After rebounding from the wall, ball B again collides with ball A. Given that ball B is brought to rest by this second collision with A,

a) show that 2e^3 - 3e^2 - 3e + 2 = 0

where e is the coefficient of restitution throughout the collisions.

b) verify that e=1/2 is the only practical solution of the equation.

If you solve this you are immense at maths lol

You have two momentum equations, two restitution equations.

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.

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.

SsEe

You have two momentum equations, two restitution equations.

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.

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

nota bene

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

A seems painful though...

A seems painful though...

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

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!!!

PLEASE HELP ME!!!

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.

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.

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.

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.

DFranklin

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

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

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

thanks mate thats great

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.

haha i managed to do the one after it, but took me ages... not 4 sides tho lol

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can someone please explain what principle domain is and why the answer is a not c?Maths

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