App Maths question

Patrick

If anyone is bothered to type of a solution to this.. I'd be very grateful

Two smooth spheres of masses m and 5m respectively lie on a smooth horizontal table. The spheres are projected towards each other with speeds 6u and 2u respectively. After the collision the spheres move in opposite directions. What is the minimum value of e for this event?

Boston

quiet simple, I wont solve it for you, just remember you can give your answer in terms of u and remember new relative velocity is a constant e times the old.

eden_my_ass

At a guess (its been a while since i saw an applied maths question) e > (4u/5) . I won't say how I got that (save my blushs). looking forward to being corrected.....

PHB

Stupid bloody minus sign!!!!
Read Sevs solution below

Patrick

thanks very much

Boston

PHB i'm open to correction here but doesn't e in that situation have a maximum of .5, thats using your maths not mine. and anyway cant e in general only have a maxium value of 1?

PHB

Yeah I see you're point, I must have made a calculation mistake, which infact mattle little in applied mahts luckily for me

Boston :

Boards doesnt allow big spaces it acctually should say

2 + 3v2/4u = e but it doesnt
I still cant work out whats wrong though

Once you can use

PCM and NEL then you can easily get close to full marks in that question.

p.s I couldnt be arsed to go back and sort out the calculation mistake

Sev

e... the coefficient of elasticity right? how can that be greater than 1 Dave? that would mean that the 2 objects bounce off each other at twice the speed at which they struck in the first place, which kind of completely disregards the notion of energy conservation.

Stop breaking down the laws of physics dave, youre slipping.

For those of you who dont know, Dave = PHB

Boston

I dont know how many marks you would get for that, I'd only give it 30 out of 50 since you made several errors and even if you hadn't made any the way your working it out will give you the maximun valley of e. In fareness i have to wonder if thats what was requested in the first place. Btw in case your wondering, you but a minus in where you should, you neglected a minus where there should have been one and you divided by 4 twice and then forgot you diveded at all. Still it not as easy as i first thought it was and i still havent been able to work out the right answer, thought it is late.

Boston

Originally posted by PHB

2 + 3v2/4u = e but it doesnt
I still cant work out whats wrong though

well you taking v2 or u to be equal to zero and neither of them are according to the question at least. It the last few minutes ive tired it several other ways and i still can't caculat a minimum value of e. probable something really simple but im to tired to notice

Sev

Right... essentially, this is how it works.

e = coefficient of elasticity

such that v2 - v1 = -e(u2 - u1)

The different in velocities after collision is equal to '-e' times the difference in their velocities before collision. So if for example 2 bodies were going at +30m/s and +20/ms before collision, and the value of 'e' was 0.5. Then you know that after collision, the difference in their velocities will be -5m/s. The trouble is, thats all you know with the elasticity rule alone...

You apply the conservation of momentum rule to find out then which then is moving at what speed after the collision.

m1(u1) + m2(u2) = m1(v1) + m2(v2)

---

so.. here we go

u1 = 6u
u2 = -2u

v1 = ?
v2 = ?

e = ?

m1 = 1m
m2 = 5m

---

Newton's experimental law of restitution

=> v2 - v1 = -e(-8u)
=> v1 = v2 - e(8u) [1]

---

Conservation of Momentum

=> (1m)(6u) + (5m)(-2u) = (1m)(v1) + (5m)(v2)
=> -4u = v1 + 5(v2)
=> v1 = -5(v2) - 4u [2]

---

Equate [1] and [2]

=> v2 - e(8u) = -5(v2) - 4u
=> 6(v2) = e(8u) - 4u
=> 6(v2) = u(8e-4)
=> v2 = (2u/3)(2e-1) [This is our expression for body 2's velocity after collision]

Likewise (with a little more playing), an expression for v1 can be found too

=> v1 = (-u/6)(10e + 1)

---

The trick is now, that for both to be moving in opposite directions to which they came after collision, v1 must be less than 0, and v2 must be greater than 0.

For v1 to be less than 0...

=> 10e + 1 > 0
=> e > -1/10

'e' can never be a negative number, so we can disregard that, Ball one will always bounce back the other direction regardless of what 'e' is. It makes sense too, because ball 2's momentum outweighs ball1's, ball1 has no hope of ploughing through much heavier ball2.

But for Ball number 2, it may or may not bounce backwards the way it came (v2>0) depending on the value of e.

So, for v2 to be greater than 0...

=> 2e - 1 > 0
=> e > 1/2

Therefore minimum of value of e is 1/2.

Sev

If you have any more, put them up... this is the only way I get any study/practice done.

Boston

Originally posted by Sev

The trick is now, that for both to be moving in opposite directions to which they came after collision, v1 must be less than 0, and v2 must be greater than 0.

For v1 to be less than 0...

=> 10e + 1 > 0
=> e > -1/10

'e' can never be a negative number, so we can disregard that, Ball one will always bounce back the other direction regardless of what 'e' is. It makes sense too, because ball 2's momentum outweighs ball1's, ball1 has no hope of ploughing through much heavier ball2.

But for Ball number 2, it may or may not bounce backwards the way it came (v2>0) depending on the value of e.

I was completely asleep last night, i got e>.5 several times and kept saying no i need the minimum value not the max. You will get the answer much quicker by putting v1 = 0 and then putting v2 = zero.

anyway

2 smooth spheres,each of mass m and radius r,collide while moving one a smooth plane. before coliltion their sppeds are 2u an u, 2u being the velocity of the sphere behind the first one. their centres move in a parallel line which are a distance 6r/5 apart. e = .5. determine the resultant velocity of each sphere and show that the angle between their paths will be 27 degrees approx.

first parts a doddle, but second part i've never been able to slove

Patrick

Sev you've got it. Thanks alot

I noticed PHB's was wrong since;

0<e<1

and from:

2u + 3v2 = e

=> e < 1/2 IIRC


What do NEL and PCM stand for btw?

Sev

Originally posted by Boston
2 smooth spheres,each of mass m and radius r,collide while moving one a smooth plane. before coliltion their sppeds are 2u an u, 2u being the velocity of the sphere behind the first one. their centres move in a parallel line which are a distance 6r/5 apart. e = .5. determine the resultant velocity of each sphere and show that the angle between their paths will be 27 degrees approx.

Well.. what I would do firstly, is rotate the image around this way.



You should be able to make that right triangle in green, from the numbers given, hypotenuse 2r in length, one side 6/5r, which means the other side is 8/5r. Quite conveniently it makes one of them neat 3:4:5 ratio triangles.

So using them ratio's of sides, I break the 2 speeds of the particles, 2u, and u, into components parallel to the line of centers of the 2 circles, and components perpendicular (which will not be affected by the collision).

Then I let p and q be vectors representing their movement after collision, and each are broken up into component vectors again, a and b, and c and d.

Already, we know that a = 6u/5, and c = 3u/5. Because my i components wont be affected by the collision.

Using conservation of momentum in the plane parallel to the line of centers.

We get..

=> (u1)(m1) + (u2)(m2) = (v1)(m1) + (v2)(m2)
=> (8u/5)m + (4u/5)m = bm + dm
=> 12u/5 = b + d [1]

Using nel again along line of centers, we get.

=> -e(u1-u2) = v1-v2
=> (-0.5)(8u/5 - 4u/5) = b - d
=> 2u/5 = d - b [2]

Add [1] and [2] together.

=> 2d = 14u/5
=> d = 7u/5

so.. b = u

so now we have

=> p = ai + bj = 6u/5i + uj = 1.56u E 36.8 Degrees N
=> q = ci + dj = 3u/5i + 7u/5j = 1.52u E 66.8 Degrees N

66.8 - 39.8 = 27 Degrees