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

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SUMMARY

The discussion focuses on solving a physics problem involving the collision of two hockey players, each weighing 75.0 kg and skating at 5.90 m/s, with an angle of 135° between their initial directions. The solution requires breaking down their velocities into horizontal and vertical components and applying the conservation of momentum in both directions. Additionally, the discussion touches on estimating the kinetic energy of a nail struck by a hammer, emphasizing the importance of using the kinetic energy formula KE = 0.5mv² and the principle of conservation of energy in elastic collisions.

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  • Understanding of momentum conservation principles
  • Knowledge of kinetic energy calculations using KE = 0.5mv²
  • Ability to resolve vectors into components
  • Familiarity with elastic collision concepts
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  • Learn about elastic and inelastic collisions in detail
  • Explore advanced momentum conservation problems
  • Investigate the effects of angles on momentum and energy transfer
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Students studying physics, particularly those focusing on mechanics, collision problems, and energy conservation principles.

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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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