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Find speed of two objects after collision

by JJones_86
Tags: collision, objects, speed
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JJones_86
#1
Sep21-08, 02:23 PM
P: 72
1. The problem statement, all variables and given/known data
Puck A has a mass of 0.226 kg and is moving along the x axis with a velocity of 5.59 m/s. It makes a collision with puck B, which has a mass of 0.452 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of

a) Puck A

b) Puck B

2. Relevant equations
No idea



3. The attempt at a solution
No Idea


------------
Where do I even start?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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G01
#2
Sep21-08, 03:30 PM
HW Helper
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P: 2,685
You have to show some work in order to get help here. Those are the rules. "No idea" doesn't cut it, sorry.

You have to have SOME thoughts on the problem. What have you tried? What concepts does the problem involve?
JJones_86
#3
Sep21-08, 04:10 PM
P: 72
Quote Quote by G01 View Post
You have to show some work in order to get help here. Those are the rules. "No idea" doesn't cut it, sorry.

You have to have SOME thoughts on the problem. What have you tried? What concepts does the problem involve?
Well I do have some thoughts...
I know that it has to do with momentum, and directions.

I really just don't know where to start.. After I find the momentum, where would I go from there?

catkin
#4
Sep21-08, 04:37 PM
P: 218
Find speed of two objects after collision

Consider the relationship between the momenta before and after collision ...
JJones_86
#5
Sep21-08, 04:59 PM
P: 72
Quote Quote by catkin View Post
Consider the relationship between the momenta before and after collision ...
Will the both have the same momentum?
catkin
#6
Sep21-08, 05:37 PM
P: 218
That's what the law of conservation tells us ... and there are more than two (you wrote "both") momenta to consider.
JJones_86
#7
Sep21-08, 05:41 PM
P: 72
Quote Quote by catkin View Post
That's what the law of conservation tells us ... and there are more than two (you wrote "both") momenta to consider.
Ok, so how would you figure out what the momentum of the two are after the collision?
catkin
#8
Sep21-08, 05:50 PM
P: 218
p = mv

Both p (momentum) and v (velocity) are vectors so it's a vector equation.

Start with the horizontal momenta, one for A before and one each for A and B after. After must be the same as before (conservation!). You don't know the magnitude of the velocities of A and B after so you'll have to call them Va and Vb (or some such) and find another equation to find their values.

I've got to go now.
JJones_86
#9
Sep21-08, 06:52 PM
P: 72
Quote Quote by catkin View Post
p = mv

Both p (momentum) and v (velocity) are vectors so it's a vector equation.

Start with the horizontal momenta, one for A before and one each for A and B after. After must be the same as before (conservation!). You don't know the magnitude of the velocities of A and B after so you'll have to call them Va and Vb (or some such) and find another equation to find their values.

I've got to go now.
Ok, so how do you find out the velocity of A & B?

So far I've come up with these equations:

P=m*v
= 0.226kg * 5.59m/s
= 1.263314 kg*m/s

Vay = Va sin 37 -- for A's Y direction
Vax = Va cos 65 -- for A's X direction
Vbx = Vb cos 37 -- for B's X direction
Vby = Vb sin 65 -- For B's Y Direction

So Va = Sqrt((Vax)^2+(Vay)^2)

Just lost on how to find the velocity of Va and Vb...
Sakha
#10
Sep21-08, 07:18 PM
P: 295
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59
JJones_86
#11
Sep21-08, 07:39 PM
P: 72
Quote Quote by Sakha View Post
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59
That just tells me the Vb is 0...
catkin
#12
Sep22-08, 01:30 PM
P: 218
Quote Quote by JJones_86 View Post
P=m*v
= 0.226kg * 5.59m/s
= 1.263314 kg*m/s
Correct.

Quote Quote by JJones_86 View Post
Vay = Va sin 37 -- for A's Y direction
Vax = Va cos 65 -- for A's X direction
Vbx = Vb cos 37 -- for B's X direction
Vby = Vb sin 65 -- For B's Y Direction
Also correct. The suffixes a, b, x and y are a smart move. Helpful to use something like 1 for before the collision and 2 for after.

Quote Quote by JJones_86 View Post
So Va = Sqrt((Vax)^2+(Vay)^2)
Correct but not useful in this problem.

Let M stand for the mass of A, 0.226 kg. How many M is the mass of B?

Slightly modifying what you wrote above, the total momentum in the x direction before the collision can be written
P1x = (V1ax * m1) + (V1bx * mb)
= (5.59 * M) + (0 * 2M)
= 5.59M

Use your expressions for velocities after the collision (V2ax etc.) in an expression for the total momentum in the x direction after the collision:
P2x = ...

What is the total momentum in the y direction before and after the collision?
P1y = ...
P2y = ...
catkin
#13
Sep22-08, 01:38 PM
P: 218
Quote Quote by Sakha View Post
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59
That's only correct if the + sign indicates vector addition.


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