Solving Elastic Collisions - Momentum & Final Velocity

AI Thread Summary
In an elastic collision involving two masses, the conservation of momentum and kinetic energy must be applied to determine the final velocities. The initial momentum for mass 1 (8 kg at 3 m/s) is 24 kg·m/s, while mass 2 (4 kg at -3 m/s) has a momentum of -12 kg·m/s. To find the final velocities, two equations are needed: one from the conservation of momentum and another from the conservation of kinetic energy. The momentum equation combines the initial and final momenta of both masses, while the kinetic energy equation ensures that the total kinetic energy before and after the collision remains constant. Solving these equations simultaneously will yield the final velocities for both masses after the collision.
bjah
Messages
3
Reaction score
0
Please clear up this problem...

Mass 1 = 8 kg, and v = 3 m/s to the right.
Mass 2 = 4 kg, and v = -3 m/s to the left.

Both objects are on the same x-plane. Totally elastic collision.
Momentum of mass 1 is 24, and momentum of mass 2 is -12.

How do I determine final momentum for each object, and therefore final velocity, after the collision?

Thank you.
Brad
 
Physics news on Phys.org
bjah said:
Please clear up this problem...

Mass 1 = 8 kg, and v = 3 m/s to the right.
Mass 2 = 4 kg, and v = -3 m/s to the left.

Both objects are on the same x-plane. Totally elastic collision.
Momentum of mass 1 is 24, and momentum of mass 2 is -12.

How do I determine final momentum for each object, and therefore final velocity, after the collision?

Thank you.
Brad

You have two unknowns (the velocities of the two masses after the collision).
If you can find two equations in these two variables, just basic algebra will see you home.

Ask yourself what the definition of an "elastic" collision is, and that will give you one of your equations. Conservation of momentum will give you the other one.
 
And that's where I'm stuck!

(8 x 3) + (4 x -3) = (8 x Vfinal of mass 1) + (4 x Vfinal of mass 2)

But how do I combine the equations for conservation of momentum and kinetic energy?
 
bjah said:
And that's where I'm stuck!

(8 x 3) + (4 x -3) = (8 x Vfinal of mass 1) + (4 x Vfinal of mass 2)

But how do I combine the equations for conservation of momentum and kinetic energy?

What does the equation for conservation of energy say about the energy before and the collision?
 
KE initial must = KE final.
 
bjah said:
KE initial must = KE final.
Right. So write out expressions for those two energies in this context.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top