Yahoo Answers: Answers and Comments for Collisions in two dimensions. Physics help!? [Physics]
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From Anonymous
enGB
Mon, 30 Mar 2020 02:36:19 +0000
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Yahoo Answers: Answers and Comments for Collisions in two dimensions. Physics help!? [Physics]
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From Ash: Lets say the puck has mass 'm' kg
Col...
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Mon, 30 Mar 2020 03:01:03 +0000
Lets say the puck has mass 'm' kg
Collision in +xdirection
m(Uax) + m(Ubx) = m(Vax) + m(Vbx)
Divide m on both sides
Uax + Ubx = Vax + Vbx
2.0 + 0 = 1.0 cos60 + Vbx
Vbx = 2.0  1.0 cos60
Vbx = 1.5 m/s
Collision in +ydirection
m(Uay) + m(Uby) = m(Vay) + m(Vby)
Divide m on both sides
Uay + Uby = Vay + Vby
0 + 0 = 1.0 sin60 + Vby
Vby =  1.0 sin60
Vby =  0.87 m/s
speed of puck B = √(Vbx² + Vby²) = √[(1.5)² + ( 0.87)²] = 1.7 m/s
direction of puck B = tan⁻¹(Vby/Vbx) =tan⁻¹( 0.87/1.5) = 30°
The direction of puck is 30° below +xdirection

From Fireman: Let the mass of the pucks is m kg and the Puck...
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Mon, 30 Mar 2020 03:16:28 +0000
Let the mass of the pucks is m kg and the Puck B velocity after the collision is v m/s at an angle θ from Xaxis.
The initial momentum of pucks on Xaxis:
Puck A: m x 2 = 2m kgm/s
Puck B: m x 0 = 0
The initial momentum of pucks on Yaxis:
Puck A: m x 0 = 0
Puck B: m x 0 = 0
The Final momentum of pucks on Xaxis:
Puck A: m x 1 x cos60* = 0.5m kgm/s
Puck B: m x v x cosθ = mvcosθ kgm/s
The Final momentum of pucks on Yaxis:
Puck A: m x 1 x sin60* = 0.87m kgm/s
Puck B: m x v x sinθ = mvsinθ kgm/s
BY the law of momentum conservation:
ON XAXIS:
=>m1u1 + m2u2 = m1v1 + m2v2
=>2m + 0 = 0.5m + mvcosθ
=>vcosθ = 1.5 (i)ON YAXIS:
=>m1u1 + m2u2 = m1v1 + m2v2
=>0 + 0 = 0.87m + mvsinθ
=>vsinθ = 0.87 (ii)
BY (ii)/(i):
=>tanθ = 0.87/1.5 = 0.58 = tan30* (below Xaxis)
=>θ = 30* (below Xaxis)
& By (i)^2 + (ii)^2
=>v^2 (sin^2θ + cos^2θ) = (1.5)^2 + (0.87)^2
=>v = √3
=>v = 1.73 m/s

From Joseph: You don't need the mass of the pucks. The...
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Mon, 30 Mar 2020 03:07:33 +0000
You don't need the mass of the pucks. They cancel out.