A classical mechanics problem by Abhinav Jha

tomorrow is my chem exam , yet doing this ! solved it like prakhar but just to give another approach !

Its a nice one.

Firstly we will assume Oblique collision.

Deutron has mass 2m and proton has mass m .

2m is moving horizontally with a velocity u and the line joining centres of the two make an angle of x with the horizontal.

Now we know that Impulse on both will act along the line of impact (line joining their centres).

So velocity of 2m perpendicular to line of impact wont change and m will get a velocity only along the line of impact.

So Assume that 2m has a velocity of v along the line of impact and m has a velocity v' along the line of impact.

obviously velocity of 2m perpendicular to line of impact will be usinx .

Now using Momentum conservation along the line of impact

2mv+mv' = 2mucosx

2v+v' = 2ucosx

Now as collision is elastic , using coefficient of restitution equation along the line of impact

v'-v = ucosx

Solving we get v = ucosx/3 , v' = 4ucosx/3

so velocity of 2m will be ucosx/3 along line of of impact and usinx perpendicular to it.

So vertical velocity of 2m =2usinxcosx/3

horizontal velocity of 2m = usin^2(x)+ucos^2(x)/3

Now let 2m is deviated by an angle theta then

tan(theta) = vertical velocity of 2m/horizontalvelocity of 2m

Differentiate wrt x to get cos2x = 1/2 or x = 30

back substitute to get theta = 30