Pool ball collision, kinetic energy

AI Thread Summary
The discussion centers on a hypothetical collision between two pool balls, where one ball strikes a stationary one, resulting in specific angles and speeds calculated using conservation of momentum. The initial kinetic energy was calculated to be 0.531 J, while the final energy was 0.334 J, indicating a loss of kinetic energy. The question raised concerns the nature of the collision's elasticity, with the conclusion that the setup implies an inelastic collision rather than a perfectly elastic one. It was noted that real-life pool ball collisions would not match the calculated angles, as perfectly elastic collisions would result in a 90-degree angle between the final velocities. Overall, the scenario illustrates the complexities of collision dynamics in physics.
krausr79
Messages
40
Reaction score
0

Homework Statement


I cooked up an example of a collision between 2 pool balls: one comes in from the right at 2.5 m/s and hits a stationary one. The original goes off at 30° up, 1m/s. What does the lower one do? I calculated the lower one to go down -17°, 1.71 m/s using conservation of momentum/ simultaneous equations. Then I calculated kinetic energies to be .531 before and .334 after (mass of pool balls assumed at .17kg)

My question is about elasticity. I didn't make any assumptions about it, so I expected to get a perfectly elastic collision (kinetic energy preserved). I've also read that pool balls are mostly elastic. Does this mean that my problem setup assumes an inelastic collision? Would it be impossible to get a pool ball collision like this in real life?


Homework Equations


pstart = pfinal
KE = 1/2MV2


The Attempt at a Solution

 
Physics news on Phys.org
Hi krausr79,

krausr79 said:

Homework Statement


I cooked up an example of a collision between 2 pool balls: one comes in from the right at 2.5 m/s and hits a stationary one. The original goes off at 30° up, 1m/s. What does the lower one do? I calculated the lower one to go down -17°, 1.71 m/s using conservation of momentum/ simultaneous equations. Then I calculated kinetic energies to be .531 before and .334 after (mass of pool balls assumed at .17kg)

My question is about elasticity. I didn't make any assumptions about it, so I expected to get a perfectly elastic collision (kinetic energy preserved).

A perfectly elastic collision is a special case, so you would need to constrain your problem for it to be elastic.


I've also read that pool balls are mostly elastic. Does this mean that my problem setup assumes an inelastic collision?

Yes.

Would it be impossible to get a pool ball collision like this in real life?

If these identical balls were assumed to be particles colliding elastically (with one initially at rest), the angle between the final velocities would be ninety degrees; this is very different from the 47 degrees that you found in your problem.
 
Ok, thanks.
 
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