Two identical pucks are on an air table. Puck A has an initial velocity of 2.0m/s in the +x-direction. Puck B is at rest. Puck A collides elastically with Puck B and A moves off at 1.0m/s at an angle of 60 degrees above the x-axis. What is the speed and direction of puck B after the collision?

Answer 1

You have two unknowns: the final x and y components of the velocity of Puck B: Vbx and Vby. The two momentum equations (x and y) will let you solve for each. You should not have to use the energy conservation equation at all; they have provided more information than you need. Mathematically speakking, the problem is overdetermined

M*Vo = M*Va*cos60 + M*Vbx
0 = M*Va*sin60 -M*Vby

2 = 1/2 + Vbx

Vbx = 3/2

Vby = sqrt3/2

Vbx^2 + Vby^2 = 9/4 + 3/4 = 3

After collision
Vb^2 + Va^2 = 4 = Vo^2
so kinetic energy really is conserved.

Answer 2

What kind of drug is this guy on. How would anyone understand that answer

Answer 3

You literally explained nothing. Your numbers just spill onto the page with no explanation of where they came from

Answer 4

Given:
Ma = M kg, Va = 2 m/s.
Mb = M kg, Vb = 0.
V3 = 1m/s[60o] CCW = Velocity of Ma after collision.
V4 = ? = Velocity of Mb after collision.

Momentum before = Momentum after.
Ma*Va + Mb*Vb = Ma*V3 + Mb*V4.
M*2 + M*0 = M*1[60] + M*V4,
2M = M[60] + M*V4,
Divide by M:

2 = 1[60] + V4,
2 = 0.5 + 0.866i + V4,
V4 = 1.5 - 0.866i = 1.732[-30o] = 30o S. of W. = 1.732m/s[210o] CCW. = Velocity of Mb after collision.

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