Dropping a ball onto a wedge, 2d elastic collisions.

[gutupio]

Homework Statement

A ball is dropped straight down onto a wedge that is sitting on top of a frictionless surface. The wedge is at an angle of 45degreees and has a mass of 5.00kg. The ball has a mass of 3.00kg.
If the ball collides with the wedge at 4.50m/s, and the collision is instantaneous and totally elastic, what are the velocities of the wedge and ball afterwards?

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Homework Equations

P1i + P2i = P1f + P2f
K1i + K2i = K1f = K2f

The Attempt at a Solution

I know the wedge, after the collision, is going to be sliding along the x-axis.
I assume because the normal force is perpendicular to the surface of the wedge, the ball will bounce away at the normal force, or a 45degree angle.

I'm confused because if the wedge only slides horizontally, with no component, that means all of the vertical component of the momentum is transferred to the ball, so the balls vertical momentum is equal to it's original vertical momentum. If the ball bounces away at a 45degree angle, it's vertical components are the same, which means it has a larger momentum than when it started, and this doesn't make sense to me.

 

[gutupio]

To be more clear, if i do:
m2<v2 cos(180), v2sin(180)> + m1<v1cos(45), v1sin(45)> = m1<0, 4.50>
since cos(180) is -1, and sin(180) is 0, this leads too:
m2<-v2 ,0> + m1<v1cos(45), v1sin(45)> = m1<0, 4.50>
This leads to two equations:

-v2m2 + m1v1cos(45) = 0
m1v1sin(45) = 4.50

These numbers leave me with vf,ball = 6.36m/s and vf,wedge=-2.70 m/s.
Again, I know these numbers are wrong, because the final velocity of the ball can't be more than the initial velocity.

 

[dacruick]

when the ball hits the wedge, there will be a component of the force that pushes the wedge into the ground, yielding no motion.

 

[gutupio]

This force is the vertical component of the force of the ball striking the wedge right? So it'd be m_ballgsin(45). It's also the same as the horizontal force, shich would cause the movement, at a Force of 20.81 N. If you divide by the mass of the wedge, you get 4.162m/s^2, which would be the acceleration of the box, but time of impact is not given, so I do not know the velocity after the collision?

 

[gneill]

Suppose for a moment that, because the collision is instantaneous, that the ball's angle of incidence is equal to its angle of reflection. Further suppose that the wedge and the Earth under it gained the appropriate momentum to balance momentum in the vertical direction (and because of the mass differential, negligible motion results in the vertical direction for the wedge+Earth). What would be the outcome for the horizontal motions of ball and wedge?

 

[gutupio]

If i do that, then the horizontal component of the moment becomes m1v1cos(180) + m2v2cos(180) = 0, or -m2v2=m1v1. So they have the same magnitude, just opposite directions. I solved that equation for v1=(-m2v2)/m1. Then, with conservation of K.E. I did Kball,i + Kwedge,i = Kball,f + Kwedge,f
Kwedge,i is 0, so
Kball,i = Kball,f + Kwedge,f or
.5m1(v1,i)^2 = .5m1(v1,f)^2 + .5m2(v2,f)^2
I plugged in v1 from above, di the algebra, and got

m1(v1,i)^2
--------------- = (v2,f)^2
m2(m2/m1 + 1)

Plugging in the original numbers, I got v2,f=2.14m/s
Since there momentums are equal, i then solved for v1 and got v1,f = 3.56m/s.
Since the wedge is moving in the negative direction, it's negative so:
Ball = 3.56 m/s and Wedge = -2.14m/s.

Is this correct?

 

[gneill]

Well, given the assumptions I made, that would be the values I would get for the velocities.

 

[gutupio]

Thank you so much, I've been wracking my brain over this problem for 5 days. It feels so good to have an answer.