collision

markosheehan · 2017-04-21 02:19:05

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.

Bob · 2017-04-21 19:46:01

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.

markosheehan · 2017-04-22 02:46:20

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??

markosheehan · 2017-04-22 03:11:19

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

Bob · 2017-04-22 11:03:13

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

markosheehan · 2017-04-24 08:25:29

are you sure.

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

Bob · 2017-04-24 09:33:12

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

markosheehan · 2017-04-24 19:09:45

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?

Bob · 2017-04-24 19:45:34

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

markosheehan · 2017-04-25 05:51:22

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

Bob · 2017-04-26 20:04:09

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

Math Is Fun Forum

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

collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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

Re: collision

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