Marble collisions, momentum, velocity

AI Thread Summary
The discussion revolves around calculating the momentum of two marbles before and after a collision. Marble A, with a mass of 5g and an initial speed of 20 cm/s, collides with marble B, which has a mass of 10g and an initial speed of 10 cm/s. After the collision, marble A moves at 8 cm/s, and participants are tasked with determining the momentum of each marble pre- and post-collision, as well as the final speed of marble B. The calculations reveal that the total momentum before the collision must equal the total momentum after, leading to an equation to solve for marble B's final speed. The key takeaway is the importance of momentum conservation in collision problems.
KatieLynn
Messages
64
Reaction score
0

Homework Statement



Marble A, mass 5g moves at a speed of 20 cm/s. It collides with a second marble B, mass 10 g, moving at 10 cm/s in the same direction. After the collision, marble A continues with a speed of 8 cm/s in the same direction. A) Calculate the momentum of each marble before the collision. B) Calculate the momentum of each marble after the collision. C) What is the speed of marble B after the collision?

Homework Equations



F=m(Vf-Vi)/t


The Attempt at a Solution



I converted all the numbers first

Mass A= .005kg
Vi A = .2 m/s
Mass B= .1 m/s
Vi B = .2 m/s
V after = .8 m/s

A) Marble A:
(.005kg)(02m/s) = .001 kg*m/s

Marble B:
(.01kg)(.1m/s)=.001 kg*m/s

B) (.001+.001)=(.005kg+.01kg)V
v=1.33 m/s

What did I do wrong in B? I'm pretty sure A is right.
 
Physics news on Phys.org
Just skimming over,

Total Momentum After = Total Momentum Before

P2' = P2

m2v2' = m2v2
(.01 Kg)(.08 m/s) = (.01 Kg)v2

algebra to isolate v2do you see what i did? Momentum after must equal momentum before, you know the momentum before the collision was m2v2, it has to equal momentum after, which is m2v2' your only unknown is v2' which you can solve using your data.. i did this usuing your definitions for the mass and velocities, i didn't check those.
 
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