Question

The drawing shows a collision between two pucks on an air-hockey
table. Puck A has a mass of 0.20 kg and is moving along the
*x* axis with a velocity of 6.60 m/s. It makes a collision
with puck B, which has a mass of 0.40 kg and is initially at rest.
After the collision, the two pucks fly apart with angles as shown
in the drawing (α = 62° and β = 40°). Find the final speed of puck
B.

Answer #1

Ma .20 kg and 6.6 m/s , 62 deg +Y, cos = .4695, sin = .8829

Mb .40 kg and 0.0 m/s , 40 deg -Y ,cos = .7660, sin = .6428

Conserving momentum along x:

Ma*Vai + Mb*Vbi = Ma*Va*cos62 + Mb*Vb*cos40

1.32 = .09389Va + 0.3064Vb _______(1)

Conserving momentum along x:

0 = Ma*Va*sin62 - Mb*Vb*sin40

0 = .1766Va - .02571Vb

Va = 1.456Vb (sub into 1)

(1) 1.32 = .09389(1.456Vb) + 0.3064Vb

1.32 = 0.1367Vb + 0.3064Vb

Vb = **2.98
m/s** (ans)

[In case you need final speed of puck A: Va = 1.456 x 2.98 =
**4.34 m/s]**

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