Calculating Final Speed in Elastic collision (momentum)

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To determine the final velocities of two colliding balls in an elastic collision, apply the conservation of momentum and conservation of kinetic energy principles. The equation m*v1 + M*u1 = m*v2 + M*u2 represents momentum conservation, while a second equation for kinetic energy conservation is also necessary. Given the masses and initial velocities, substitute the known values into these equations to solve for the unknown final velocities. It's crucial to remember that both momentum and energy are conserved in elastic collisions. This approach will yield the final speeds after the collision.
Mushroom79
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Homework Statement



If it is eg. two balls, both going to the right before and after an elastic shock, how do you determine their velocities after the collision? (speeds before and weights are given)

Eg.

M = 50 kg
u1 = 20 m / sm = 100 kg
v1 = 10 m / s

(M, m = mass)
(u1, v1 = velocity before the collision)
(u2, v2 = velocity after collision)

Homework Equations



The law of conservation of momentum:
m*v1+M*u1= m+M*v2*u2

Momentum before = Momentum after

The Attempt at a Solution



m*v1+M*u1= m+M*v2*u2 →

100 * 10 +50 * 20 = 100 +50 * v2 * u2

How do I continue from here?
 
Last edited:
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In elastic collisions energy is also conserved , so you can set up one more equation .
 
ZxcvbnM2000 said:
In elastic collisions energy is also conserved , so you can set up one more equation .

Oh, right. Think "momentum before equals momentum after" is what you meant.
Forgot to put it there.
 
In elastic collisions :

Momentum is conserved

Energy Is conserved
In inelastic collisions:

Momentum Is conserved

Energy is not conserved

Just to be more clear :P
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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