Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2017-04-21 02:19:05

markosheehan
Member
Registered: 2016-06-15
Posts: 51

collision

Two smooth spheres of masses km and m collide obliquely. the sphere of mass m is brought to rest by the impact. the coefficient of restitution for the collision is 1/k (k greater or equal to 1) Prove before the impact the spheres were moving perpendicular to each other.

i have worked out k=-1. I know the sphere of mass m was xi+0j before and 0i+0j after.

Offline

#2 2017-04-21 19:46:01

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,969

Re: collision

hi markosheehan

You can use a similar approach to the previous one.  Take x as along the line of centres. I used Wx and Wy  for km before ; Ux and Uy  for m before ; Vx and Vy for km after.

You can make 4 equations: conservation of momentum in each direction; experimental law in the x direction; no change of relative velocity in the y direction*.

This last will allow you to show that Uy is zero.  Looks like you already have Wx is zero.

Bob

* These problems are the first time I've had to use this 'no change' law, but I think it makes sense in terms of Newton's laws.  In the y direction the spheres just scrape past each other without any 'bounce' so why should their relative velocities change.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#3 2017-04-22 02:46:20

markosheehan
Member
Registered: 2016-06-15
Posts: 51

Re: collision

i have used conservation of momentum in only the i direction as its pointless using it in the j direction.
a(km)+x(m)=y(km)+0(m)
then using the coefficient     or restitution formula  y-o/a-x =-1/k

solving these k=-1

actually wait i think i have it.
so when we compare these equations   -a+x=yk   and ak+x=yk they are inconsistent unless a=0  that means initially there speeds are bj for the sphere of mass km and xi for the sphere of mass m
Is this right??

Offline

#4 2017-04-22 03:11:19

markosheehan
Member
Registered: 2016-06-15
Posts: 51

Re: collision

the second part of the question is show that as a result of the collision the kinetic energy lost by the sphere of mass m is k times the kinetic energy gained by the sphere of mass km.
the kinetic energy lost by mass m sphere is .5mx^2
kinetic energy gained by mass km sphere is .5(km)(y)^2-.5(km)(a)^2

Offline

#5 2017-04-22 11:03:13

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,969

Re: collision

hi markosheehan

I agree that a=0.  But you'll have to consider y direction too if you want to complete this question,

I don't see where you got that value of k from.  It must be wrong as k cannot be negative.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#6 2017-04-24 08:25:29

markosheehan
Member
Registered: 2016-06-15
Posts: 51

Re: collision

are you sure.

i was thought the coefficient of friction always has to be negative

Offline

#7 2017-04-24 09:33:12

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,969

Re: collision

hi markosheehan

The coefficient of restitution is taken to be a positive number between 0 and 1 (inclusive).  For an object that bounces off a fixed surface, clearly it will change direction after the impact.  For a pair of objects colliding and bouncing apart it is necessary to consider the relative velocities.  The law, usually attributed to Newton, incorporates a negative sign in order to cover this point:

Note that, in the question, positivity is implied:

the coefficient of restitution for the collision is 1/k (k greater or equal to 1

I have successfully done all parts of this question.  I started with initial velocities for each as two components in the direction made by the centres (x) and perpendicular (y).  Similarly two components after the collision.

I constructed four equations: (1) conservation of momentum in the x direction; (2) same in the y direction; (3) restitution law in the x direction; (4) no change in relative velocity in the y direction.

I don't think you will succeed with less.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#8 2017-04-24 19:09:45

markosheehan
Member
Registered: 2016-06-15
Posts: 51

Re: collision

Ok i think i have the four equations. I will post them
Kma+mx=ykm
Kmb=kmb
a-x/y =-1/k
B=b

A stands for the initial velocity of the km sphere in the i direction. B stands for the initial and final velocity of the sphere km in the j direction. Y stands for the final velocity of the sphere km in the i direction. X stands for the initial velocity of the m sphere in the x direction.
These are not right though are they?

Offline

#9 2017-04-24 19:45:34

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,969

Re: collision

Initially you need 8 components for a problem of this type.  You are given that two are zero, but must prove that some others are also zero.  I suggest

mass             km                m


before            Ai                 Xi
                     Bj                 Wj

after               Yi                  0
                      Cj                 0

Your first equation is correct.

The second is the wrong way up.  Final relative velocity above and initial relative velocity below.
From these two you can show that A = 0

Third is     Bkm + Wm = Ckm
fourth is    B - W = C

From these you can show that W = 0 and B = C

A = W = 0 shows that the initial velocities are perpendicular.

B = C shows that the j component velocity for km is unchanged by the impact.

Now you can calculate the changes in KE.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#10 2017-04-25 05:51:22

markosheehan
Member
Registered: 2016-06-15
Posts: 51

Re: collision

i am not sure how your get equation 3 and 4.
using your letters:
the second equation is bkm+wm=ckm i thought and not the third equation?
now when i try to get the third equation i use the formula v1-v2/u1-u2 =-e  so   a-x/y-0 =-1/k
for the fourth equation i get c=b. i am not sure where you get b-w=c

when i solve these i get w=0
now to prove the sphere of mass km is moving perpendicular
i can rearrange  a-x/y-0 =-1/k to ka-kx=-y  now when i compare this formula with equation 1 they should be similar. the only way to make them similar is if a=0
so thats the first part of the question.
for the second part i am still confused

here is another similar question

Two smooth spheres of masses 4 kg and 2 kg impinge obliquely.
The 2 kg mass is brought to rest by the impact. 
(i)    Prove that, before impact, they were moving in directions perpendicular to each other.
(ii)    Show that, as a result of impact, the kinetic energy gained by the 4 kg mass is equal to half that lost by the 2 kg mass.

the worked answer to the question is this
http://thephysicsteacher.ie/Exam%20Material/AppliedMaths/Scans/5.%20Collisions/1986%20b.pdf
for some reason in the second part of the question he only takes into consideration the i direction which does not make sense to me

Offline

#11 2017-04-26 20:04:09

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,969

Re: collision

First question  What have you got for the change in KE?


next question: Looks like the first with values for the velocities.

he only takes into consideration the i direction which does not make sense to me

for m you have shown that W = 0 so there is no change of KE in the y direction ( zero to zero)

For Km you know C = B so again no change in KE in the y direction.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

Board footer

Powered by FluxBB