Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

# Collisions in two dimensions. Physics help!?

I dont have any mass , I don’t know how to start !

Relevance
• 2 months ago

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 X-axis.

The initial momentum of pucks on X-axis:

Puck A: m x 2 = 2m kg-m/s

Puck B: m x 0 = 0

The initial momentum of pucks on Y-axis:

Puck A: m x 0 = 0

Puck B: m x 0 = 0

The Final momentum of pucks on X-axis:

Puck A: m x 1 x cos60* = 0.5m kg-m/s

Puck B: m x  v  x cosθ  = mvcosθ  kg-m/s

The Final momentum of pucks on Y-axis:

Puck A: m x 1 x  sin60* = 0.87m kg-m/s

Puck B: m x v x sinθ = mvsinθ kg-m/s

BY the law of momentum conservation:

ON X-AXIS:

=>m1u1 + m2u2 = m1v1 + m2v2

=>2m + 0 = 0.5m + mvcosθ

=>vcosθ = 1.5 --------------------------(i)ON Y-AXIS:

=>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 X-axis)

=>θ = 30* (below X-axis)

& By (i)^2 + (ii)^2

=>v^2 (sin^2θ + cos^2θ) = (1.5)^2 + (0.87)^2

=>v = √3

=>v = 1.73 m/s

• Joseph
Lv 7
2 months ago

You don't need the mass of the pucks.  They cancel out.

• Ash
Lv 7
2 months ago

Lets say the puck has mass 'm' kg

Collision in +x-direction

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 +y-direction

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 +x-direction

• NCS
Lv 7
2 months agoReport

nice