Collision of Equal Mass Hockey Pucks

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In a collision between two equal mass hockey pucks, momentum conservation is key, not equal distribution of velocity. The initial moving puck has a speed of 5.4 m/s, while the other puck starts at rest. After the collision, the pucks move at angles of 33° and 46° to the original direction. The x-components of their velocities will not simply average to 2.7 m/s; instead, both momentum in the x and y directions must be conserved to determine their final speeds. Understanding these principles is crucial for solving the problem accurately.
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Homework Statement


Two hockey pucks of equal mass undergo a collision on a hockey rink. One puck is initially at rest, while the other is moving with a speed of 5.4 m/s. After the collision, the velocities of the picks make angles of 33° and 46° relative to the original velocity of the moving puck. Determine the speed of each puck after the collision.

I just need a confirmation of the concepts involved. Since the pucks are of equal mass, does that mean that the velocity of the first puck becomes distributed evenly between the two pucks? So would the x-components of the velocity of the pucks after the collision both be 2.7 m/s?
 
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chudzoik said:

Homework Statement


Two hockey pucks of equal mass undergo a collision on a hockey rink. One puck is initially at rest, while the other is moving with a speed of 5.4 m/s. After the collision, the velocities of the picks make angles of 33° and 46° relative to the original velocity of the moving puck. Determine the speed of each puck after the collision.

I just need a confirmation of the concepts involved. Since the pucks are of equal mass, does that mean that the velocity of the first puck becomes distributed evenly between the two pucks? So would the x-components of the velocity of the pucks after the collision both be 2.7 m/s?

Not in general. What's conserved is momentum. The total momentum in the x-direction will be conserved, as will the total momentum in the y-direction.
 
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