How Do Elastic Collisions Affect Velocities and Center of Mass Movement?

AI Thread Summary
In an elastic collision involving two balls of different masses and velocities, the conservation of momentum and kinetic energy principles are crucial for determining their final velocities. The initial momentum and kinetic energy can be calculated using the given masses and velocities. After the collision, the total momentum and energy remain constant, leading to a system of equations that can be solved for the final velocities of both balls. The center of mass velocity before and after the collision can also be calculated using the masses and initial velocities. Understanding these concepts is essential for solving the problem accurately.
DatAshe
Messages
9
Reaction score
0

Homework Statement


A ball of mass 0.206 kg with a velocity of 1.54 m/s meets a ball of mass 0.291 kg with a velocity of -0.396 m/s in a head-on, elastic collision.
(a) Find their velocities after the collision.
1f = m/s
2f = m/s
(b) Find the velocity of their center of mass before and after the collision.
cm, before = m/s
cm, after = m/s



Homework Equations


conservation of momentum? p=mv, pf=pi+I, KEf=KEi (because its elastic), Vcm=Ʃmivi/Ʃmi


The Attempt at a Solution


I have spent hours fiddling around with equations and I can't get anything to result in a correct answer. I am stumped and frustrated. This is past homework and I am just looking for an explanation on how I get from this information to the final answers. Step by step would be much appreciated, especially if you use calculus because I am lost and I am only to derivatives in calc I so far. Thank you for any and all help!
 
Physics news on Phys.org
Yes, conservation of energy and of momentum.
What is the total momentum before? Total energy before?
In terms of the unknown final velocities, what are the energy and momentum after collision?
What equations do you get?
 
wouldn't the before and after be the same do to conservation of momentum and energy?
so mivi^2=mfvf^2 and mivi=mfvm?
 
You have to calculate the total momentum and energy.

That is the momentum of both balls added together. The sum total is the same before and after the collision. m1*v1i + m2*v2i = m1*v1f + m2*v2f

Same with energy.

That gives you a couple of equations with two unknowns (v1f and v2f) to solve.
 
Last edited:
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}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top