Elastic collision/ momentum question w/ conservation of energy

In summary, a 7.0 kg rubber ball moving at 20 m/s along the x-axis collides elastically with a 25 kg ball. After the collision, the 7.0 kg ball's velocity is directed in the positive y-axis. To find the final speed of the 7 kg ball and the velocity of the 25 kg ball, the equations m1v1= m1v1f+m2v2f and 1/2m1v12= 1/2m1v1f2 +1/2m2v2f2 can be used. By setting the x momentum of the smaller mass equal to the final x momentum of the bigger mass, and using the energy
  • #1
mjmontgo
10
0

Homework Statement



In deep space a 7.0 kg rubber ball moving along the x-axis with a speed of 20 m/s collides ELASTICALLY with another ball of mass = 25 kg. After the collision the 7.0kg rubber ball's velocity is directed in the positive y axis. Find the final speed of the 7 kg ball, as well as the VELOCITY of the 25 kg ball (speed and direction)

Homework Equations


m1v1= m1v1f+m2v2f

1/2m1v12= 1/2m1v1f2 +1/2m2v2f2

The Attempt at a Solution



so i know that because the smaller mass is afterwards directed in the positive y direction, that all of its x momentum is transferred to the bigger mass. I've found a velocity for the x component of the bigger mass, however I am stucking finding the y component, which in turn would give me the over all final velocity, which i can use to calculate the angle as well as the final speed of the smaller mass. Would be greatly appreciated if someone could point out the extra equation I am missing. Basically what i have so far boils down to these equations:

m1v1x= m2v2finalx and i work out the x component of the larger mass to be 140/25 m/s

the other equation i have is for the y direction and that is :
-m2v2fy= m1v1fy

i know i somhow need to use the energy conservation formula but I am just not seeing it.
Thanks a lot in advance for you help and time guys.
 
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  • #2
You're doing fine. Write out the energy conservation using v2 = vx2+vy2
 
  • #3
Ok, so I've rewritten the energy conservation formula and what it comes down to is:

m1v1xi2= m1v1yf2+m2v2xf2+m2v2yf2


with some substitution with my known quantities i.e mass and m2v2xf=140/25

i get 2016= 7kgv1yf2+25v2yf2

which i solve in terms of v1yf and sub into -m2v2fy=m1v1yf

but when i calculate the velocity i get a negative answer, which doesn't make sense for v1yf because we already know from the question that it's moving in the positive y- direction after the collision.

Maybe I'm just making a silly arithmetic error but what i get is:

-sqrt(2016-7kg(v1yf2))=25kgv1yf2 and you can't take sqrt of a negative obviously?
 
  • #4
-m2v2fy=m1v1yf

From this momentum equation you have one velocity in terms of the other, substitute into your energy equation.
 
  • #5
Still, your method should work.
2016= 7kgv1yf2+25v2yf2 is fine. How come I continue with
2016 - 7kgv1yf2 = 25v2yf2 and never find a minus sign ?

Ah, I see, you solve in terms of v1 means you get v2,y2 = (2016 - 7kgv1yf2)/25 which has 2 solutions, right ?

And you continue with which one ?
Whatever, don't forget to check your outcome, because I think I'm missing something still.
 
Last edited:
  • #6
mjmontgo said:
with some substitution with my known quantities i.e mass and m2v2xf=140/25
You mean v2xf=140/25 m/s, yes?
i get 2016= 7kgv1yf2+25v2yf2

which i solve in terms of v1yf and sub into -m2v2fy=m1v1yf

-sqrt(2016-7kg(v1yf2))=25kgv1yf2
I don't see how you get that last equation. It doesn't make sense dimensionally. Did you mean -(2016-7kg(v1yf2))=25kgv1yf2? That would be a sign error somewhere. I suspect you squared -m1v1yf/m2 and left the result as negative.
 

Related to Elastic collision/ momentum question w/ conservation of energy

What is an elastic collision?

An elastic collision is a type of collision between two bodies in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

What is momentum?

Momentum is a measure of an object's motion. It is defined as the product of an object's mass and its velocity. In a closed system, the total momentum remains constant, meaning that the momentum of the objects before and after a collision must be equal.

How does conservation of energy apply to elastic collisions?

Conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In an elastic collision, the total kinetic energy of the system is conserved, meaning that no energy is lost during the collision. This applies to both the kinetic energy and potential energy of the objects involved.

How is the coefficient of restitution related to elastic collisions?

The coefficient of restitution is a value that represents the ratio of the relative velocities of two objects after a collision to their relative velocities before the collision. In an elastic collision, the coefficient of restitution is equal to 1, meaning that the relative velocities of the objects after the collision are equal to their relative velocities before the collision.

What factors can affect the outcome of an elastic collision?

The outcome of an elastic collision can be affected by factors such as the mass and velocity of the objects involved, any external forces acting on the objects, and the angle at which the objects collide. In an ideal elastic collision, the objects will rebound with the same speed and direction as before the collision.

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