How Does Mass Affect Velocity in an Elastic Collision?

AI Thread Summary
The discussion centers on the elastic collision between two pucks on an air-hockey table, focusing on the effects of mass and velocity. Puck A, with a mass of 0.023 kg and a velocity of +5.5 m/s, collides with puck B, which has a mass of 0.048 kg and is initially at rest. Participants are troubleshooting their calculations for momentum conservation in both the x and y directions, noting discrepancies in their equations. A key point of confusion arises regarding the correct application of initial and final velocities in the momentum equations. The conversation emphasizes the importance of correctly applying conservation laws and clarifying variable definitions in collision problems.
mayanfox
Messages
2
Reaction score
0
The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.023 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.048 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.

http://www.webassign.net/CJ/07_32.gif

(Puck A is 65 degrees north of east, puck B is 37 degrees south of east.)


I used MaVa = MaVa(cos 65) + MbVb(cos 37), which gives me
.1265 = .0097202Va + .03833Vb. Since the total momentum in the Y direction was zero, I also used
Va(sin 65) = Vb(sin 37)
Va = .6640Vb or Vb = 1.506Va.

Every time I plug in those numbers I get a wrong value! Am I missing some important concept, or am I just making an algebraic error?
 
Physics news on Phys.org
> I used MaVa = MaVa(cos 65) + MbVb(cos 37)

You Va on RHS should be different from initial Va on LHS.
 
Aah, thanks for pointing that out. I had the Va on the LHS as Vo when I did the problem, though, so it didn't change my answer.
 
> Va(sin 65) = Vb(sin 37)

Why? It should be MaVasin 65 = MbVbsin 37.
 
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