Question

A collision between two identical pucks, one moving and one stationary, takes place on ice. The puck with an initial momentum of 4.0 kg is deflected 60 degrees eastward from its original path. Using the law of conservation of momentum, what is the momentum of the puck that was originally stationary after the collision?

Here is what I understand: Assume initial momentum is on the y axis with no x component. Deflection is negative 60 degrees. Puck Stationary has 0x and 0y. Pi=Pf. Pxi=Pxf. Pyi=Pyf.

I have had this answered before and I think the answer was correct, but I still don't understand it.

First answer I received used sine when I would have used cosine and vice versa. My way produced an illogical answer so I don't think my logic is right.

To help me understand this, would you please draw when you respond? Thanks.

Answer 1

By conservation of momentum

momentum before collision = momentum after collision

let direction of moving puck is along +y axis.

conservation of momentum along y axis

(finally object is moving in a direction inclined at an angle of 60⁰ with y-axis, so resolving momentum in perpendicular directions i.e. 4cos60 along y axis and 4sin60 along x-axis, Refer diagram below)

4 + 0 = 4*cos60 + py

py = 2

conservation of momentum along x axis

0 + 0 = 4*sin60 + px

px = - 4*0.8660254 = 3.464

Momentum of 2nd puck = sqrt(2^2 + 3.464^2) = 4

theta = arctan(2/-3.464) = - 30 deg

Direction is 30 deg clockwise from -x axis or 60 deg counterclockwise from +y axis

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