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: $$[{1 \over \sqrt{2}} , {1 \over \sqrt{2}}] \\$$


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

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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

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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