Question

Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is traveling to the left, and the particle 3m is traveling to the right. They undergo an elastic glancing collision such that particles m is moving in the negative y direction after the collision at a right angle from its initial direction.
(a) Find the final speeds of the two particles in terms of Vi.

(b) What is the angle theta at which the particle 3m is scattered?

Answer 1

Answer:

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

ANGLE is 35.3 degree celcius

Explanation:

Given data:

mass m and 3m

initial speed Vi

particle with mass m is moving toward left while particle with mass 3m is moving toward right

By using conservation of momentum :

$mVi + 3m(-Vi) = mv1 +3mv2$

$-2mVi = m(v1 + 3v2)$

$-2Vi = v1 + 3v2$

conservation of energy :

$m(Vi^2) + 3m(-Vi^2) = mv1^2 + 3mv2^2$

$4mVi^2 = m(v1^2+3v2^2)$

$4Vi^2 = v1^2+3v2^2$

After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,

x direction : $-2mVi = 3m.v2i$

$-2Vi = 3v2i$

y direction : $0 = m(v1)j+3m(v2)j$

$-v1j = 3v2j$

subsitute these value in energy conservation

$v1 = \sqrt{(2)}Vi$

$v2 = \sqrt{(2/3)}Vi$

$angle = tan^{-1}(\frac{\sqrt{(2)}}{2}) =$35.3 degree from x-axis