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The problem: A car with a mass of 1200 kg and speed 10 m/s runs into a traffic barrier at an angle of 45 degrees, and is thrown outwards at an angle of 45 degrees relative to the barrier with a speed of 5.0 m/s How big was the change in momentum of the car during the collision?

I tried decomposing the momentum, but failed to get the right answer. Can someone explain to me how I can solve this question without using the fact that Pinitial and Pfinal make a right triangle?

Additionally, someone made the following solution to this problem:

$$\vec{p_e} = 1200 kg \cdot 5 m/s [{1 \over \sqrt{2}} , {1 \over \sqrt{2}}] \\ \vec{\Delta p} = \vec{p_e} - \vec{p_f} = 1200 [-{5 \over \sqrt{2}}, {15 \over \sqrt{2}}] kg m/s \\ |\vec{\Delta p}| = 1200 \sqrt{({5 \over \sqrt{2}})^2 + ({15 \over \sqrt{2}})^2} kg m/s = 1,3 \cdot 10^4 kgm/s$$

Can someone please explain to me how vectors and trig are used to solve this? I especially don't understand what the following means, and where it came from: $$[{1 \over \sqrt{2}} , {1 \over \sqrt{2}}] \\$$

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closed as off-topic by jinawee, user10851, Brandon Enright, tpg2114, Kyle Kanos Feb 21 '14 at 21:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – jinawee, Community, Brandon Enright, tpg2114, Kyle Kanos
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  • $\begingroup$ If this is a question about a previous post you shoud add a comment to that post, frist or at least link to it. $\endgroup$ – ja72 Feb 21 '14 at 19:47
  • $\begingroup$ Unfortunately the post was from a thread on a non-english high school physics site , and the post was made in 2007. I just asked it here because I don't think the OP is still on the original site. $\endgroup$ – Ali Mustafa Feb 21 '14 at 19:50
  • $\begingroup$ $[a,b]$ means a vector with length $a$ in the $x$ and length $b$ in the $y$ direction. $\endgroup$ – Kyle Kanos Feb 21 '14 at 21:29
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Momentum is vector and it has a direction. Since the motion is at 45° from the collision normal axis, the direction vector is

$$\left[ \sin 45°, \cos 45° \right] = \left[ \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right]$$

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