To marble balls colliding find the velocity after collision?

Question

Hi,

so the problem is a 30 g marble ball which moves to the right with a speed of 0.200 m/s and a 10g ball coming from the right. with a speed of 0.4 m/s. They collide and I want to find the velocity after they collide. I got this far

mv1+mv2=mv1f+mv2f
0.03*0.2+0.01*-0.4 = 0.03v1f + 0.01v2f
0.002=0.03v1f + 0.01v2f

then i used
v2f-v1f = -(v2-v10)
v2f-v1f = -.6

then i solved for v2f and substituted it into the first equation.
.002 = 0.04v1f+0.01 (-.6 +v1f)
...
v1f=.2

v2f= -.6+.2 = -.4

those final velocities are wrong they should be -0.1 and 0.5 and also i think v1f should have been negative cause it is going in the opposite direction after the collision. I'm not sure what i did wrong but i really appreciate any help.

Ok so I understand what you did and did the problem again ang got the right answers. The only thing I don't understand is why i can't use v2f-v1f = -(v2-v1)
because I found that formula in my book and on the formula sheet for my finals as Perfectly elastic Conservation of energy for 1-D.

Answer 1

*
* m1 v1+m2 v2=m1 v1f+m2 v2f
*
Correct. It's momentum conservation.


* then i used
* v2f-v1f = -(v2-v10)
* v2f-v1f = -.6
Nope.
There is no such "law of conservation of relative velocity". Lets go by the book and use conservation
of energy.
You do not say so, but I suspect that your book explicitly
states that collision is elastic.


m1 = 30g
v1= +0.2 m/s
m2 = 10g
v2 = -0.4 m/s

Conservation of momentum:
m1 v1 + m2 v2 = m1 v1f + m2 v2f
m1(v1f - v1) = -m2(v2f - v2) [1]

Conservation of energy:
1/2 m1 v1² + 1/2 m2 v2² = 1/2 m2 v1f² + 1/2 m2 v2f²
m1(v1f² - v1²) = -m2(v2f² - v2²)
m1(v1f - v1)(v1f + v1) = -m2(v2f - v2)(v2f + v2) [2]


Divide [2] by [1]
v1f + v1 = v2f + v2 <<-------this is your second equation
v1f = v2f + v2 - v1
substitude v1 into conservation of momentum [1]

m1(v1f - v1) = -m2(v2 -v2f) <<---- conservation of momentum
m1((v2f + v2 - v1) - v1) = -m2(v2f -v2)
m1 v2f + m1(v2 - 2v1) = - m2 v2f + m2 v2
(m1 - m2) v2 - 2m1 v1= - (m2 + m1) v2f


v2f = (m2 - m1)/(m1 + m2) v2 + 2m1/(m1 + m2) v1
similarily
v1f = (m1 - m2)/(m1 + m2) v1 + 2m2/(m1 + m2) v2




Answer:
v1f = 20/40 * 0.2 + 2*10/40 *(-0.4) = 0.1 - 0.2 = -0.1 m/s
v2f = -20/40 * -0.4 + 2*30/40*0.2 = 0.2 + 0.3 = 0.5 m/s

Answer 2

Ok, the first thing you did was say that momentum is conserved and you got .002 = .03v1f +.01v2f
Let's make that cleaner by multiplying both sides by 100:
.2 = 3v1f + v2f
v2f = .2 - 3v1f

That's the first equation, but I don't understand what you did next:

v2f-v1f = -(v2-v10)
v2f-v1f = -.6

What you SHOULD have done was write an equation about kinetic energy because kinetic energy will be conserved.

Before the collision, the kinetic energy is
KE1 = .5*m1*v1^2 + .5*m2*v2^2
KE1 = .5*.03*(.2)^2 + .5*.01*(-.4)^2
KE1 = .0006 + .0008 = .0014 J

After the collision:
KE2 = .5*m1*v1f^2 + .5*m2*v2f^2
KE2 = .5*.03*v1f^2 + .5*.01*v2f^2
KE2 = .015*v1f^2 + .005*v2f^2

and KE1 = KE2 so .015*v1f^2 + .005*v2f^2 = .0014

Use the momentum equation to substitute an expression for v2f into the kinetic energy equation to eliminate v2f:

.015*v1f^2 + .005*(.2 - 3v1f)^2 = .0014
.015*v1f^2 + .005*(.04 - 1.2v1f + 9v1f^2) = .0014
.015v1f^2 + .0002 - .006v1f + .045v1f^2 = .0014
.06v1f^2 - .006v1f - .0012 = 0

Let's divide both sides by .06 to clean up the equation.

v1f^2 - .1v1f - .02 = 0
factoring:
(v1f + .1)*(v1f - .2) = 0
v1f = -.1 or v1f = .2
The second answer does not make sense because that is what the initial condition was and the marble will not continue on with the same velocity after colliding with something coming from the right.

Substitute these into your momentum equation:
v2f = .2 - 3v1f
v2f = .5

Answer 3

your mistake is in the assumption of v final.
in an elastic collision between bodies of diff mass
v1f = [(m1-m2)/(m1+m2)] * v1 + [(2*m2/(m1+m2)] * v2
= [20g/40g]*0.2 + [20g/40g]* -0.4
=1/2 [0.2-0.4]
= -0.1

then you can do the same calc for v2f, or you can calc the momentum of m1, and see what momentum is left for m2, and then find it's v.