How Do Angles Affect Momentum and Kinetic Energy in Physics Problems?

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The discussion focuses on solving a physics problem involving the collision of two hockey players, emphasizing the need to analyze momentum and kinetic energy. The players, each weighing 75 kg and moving at 5.90 m/s, collide at a 135° angle, requiring the calculation of their combined speed post-collision. Participants suggest using trigonometric functions to break down the velocities into horizontal and vertical components, highlighting the conservation of momentum in both directions. Additionally, a separate problem regarding a hammer striking a nail is mentioned, with an emphasis on estimating kinetic energy and confirming that initial kinetic energy equals final kinetic energy in elastic collisions. The conversation underscores the importance of applying physics principles accurately to solve these types of problems.
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Homework Statement



Two 75.0 kg hockey players skating at 5.90 m/s collide and stick together. If the angle between their initial directions was 135°, what is their speed after the collision? (Let the motion of player 1 be in the positive x-direction and the motion of player 2 be at an angle of 135° measured counterclockwise from the positive x-axis.)

also the answer is to be the velocity in the x direction, and the velocity in the y direction

Homework Equations



I've tried a bunch relating to momentum..

The Attempt at a Solution


I tried using trig, but there must be and easier way that I'm missing


ONE MORE:
The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 11 g nail when it is struck by a 550 g hammer moving with an initial speed of 4.9 m/s.

I know that KE = .5mv^2 and that this is elastic, therefore the initial KE must equal the final KE. I got 2.4, which seems close, but still incorrect.
 
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For the hockey question, do use trig. Break the velocities for both players into vertical and horizontal components.

For the player going in the x-direction: p_x = mv

For the other player: p_x = -mvcos45 and p_y = mvsin45

Also, remember that momentum is conserved for both the horizontal and vertical directions. Vertically,

mvsin45 = (m+m)v_after.

Do the same for the horizontal component. I've probably done too much as it is.
 
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