Conservation of Momentum Problem

[skier07]

Homework Statement

(understood: two air hockey pucks make contact--the x-axis is the table)
Puck A has a mass of .025 kg and is moving along the x-axis with a velocity of +5.5m/s. It makes a collision with puck B, which has a mass of .05kg and is initially at rest. The collision is NOT head on. After the collision, the two pucks fly apart with the following angles.
Find the final speed of a) puck A and b) puck B

A:
Ma=.025kg
Voa=5.5m/s
Vfa= ?

B:
Mb=.05 kg
Vob = 0m/s
Vfb= ?
......A
...../
....../ 65 deg
-----A--->--B-------------
......\ 37 deg
.....\
......B

Homework Equations

(Ma)(Vfa) + (Mb)(Vfb) = (Ma)(Voa) + (Mb)(Vob)

Also, this equation can be written in terms of X and Y

X: (Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voax) + (Mb)(Vobx)
Y: (Ma)(Vfay) + (Mb)(Vfby) = (Ma)(Voay) + (Mb)(Voby)

The Attempt at a Solution

Since it is an air hockey table, friction is negligible.
therefore, Momentum is conserved, and Fnet=0

Initial momentum in Y = 0
Initial momentum in x = (Ma)(Voax) + 0 = .1375 kgm/sec

 

[Nate810]

Momentum in Y after collision = (Ma)(Vfay) + (Mb)(Vfby) = 0 Momentum in x after collision = (Ma)(Vfax) + (Mb)(Vfbx) = .1375 kgm/secTherefore,X: (Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voax) + 0Y: (Ma)(Vfay) + (Mb)(Vfby) = 0 Using trigonometry, Vfay = Voa * sin(65) = 3.53 m/sVfox = Voa * cos(65) = 4.19 m/s Vfby = Voa * sin(37) = 2.43 m/sVfbx = Voa * cos(37) = 4.98 m/sTherefore, Vfa = √(Vfay^2 + Vfox^2) = 5.46 m/sVfb = √(Vfby^2 + Vfbx^2) = 5.10 m/s

 

[Moetasim]

Final momentum in Y = (Ma)(Vfay) + (Mb)(Vfby) = 0
Final momentum in X = (Ma)(Vfax) + (Mb)(Vfbx) = .1375 kgm/sec

Using the given angles, we can split the final velocities into their X and Y components using trigonometry.

For puck A:
Vfax = Vfa*cos(65) = 2.39 m/s
Vfay = Vfa*sin(65) = 4.99 m/s

For puck B:
Vfbx = Vfb*cos(37) = 4.02 m/s
Vfby = Vfb*sin(37) = 2.99 m/s

Plugging these values into the conservation of momentum equation, we get:

(Ma)(Vfax) + (Mb)(Vfbx) = (Ma)(Voa) + (Mb)(Vob)
(.025 kg)(2.39 m/s) + (.05 kg)(4.02 m/s) = (.025 kg)(5.5 m/s) + (.05 kg)(0 m/s)
.06 kgm/s + .2 kgm/s = .1375 kgm/s + 0
.26 kgm/s = .1375 kgm/s

Therefore, the final speed of puck A is Vfa = .26 kgm/s and the final speed of puck B is Vfb = .1375 kgm/s.