I just need help with part b. for part a I know the answers are 0.445 and 0.894

Question

I just need help with part b. for part a I know the answers are 0.445 and 0.894

A proton, moving with a velocity of vi, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 0.50 times the speed of the proton initially at rest, find the following. (a) the speed of each proton after the collision in terms of v initially moving proton initially at rest proton Your response differs from the correct answer by more than 10%. Double check your calculations. × vi (b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction) initially moving proton nitially at rest proton o relative to the +x direction relative to the +x direction

Explanation / Answer

initially moving proton goes in the positive y direction as mentioned in the question

now using momentum balance we know that the rest proton will have velocity as x (i) - 0.445 (j)

we know that its speed is 0.894 so we find out the value of x using pythagoras : x = 0.775 x Vi

so the direction is tan-1(-0.445/0.775) = - 30 degrees

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