An elementary particle moving at 1.35 x 10^5 m/s strikes an identical particle initially at rest. After the?

Question

collision, the particles (which cannot be distinguished) are observed moving at angles of 41.7 and 48.3 degrees relative to the original direction of motion.
Find the final speed of each particle.

Answer 1

Solve for momentum.

Before collision total momentum = mv

Define your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.

particle 1 goes flying off at v1 at angle a1

particle 2 goes flying off at v2 at angle a2

The total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).

So m*v1*sin(a1) = m*v2*sin(a2)

Similarly the x components must add up so

m*v1*cos(a1) + m*v2*cos(a2) = mv

Now just solve the first equation for V1 and substitute in the second equation.

Then solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles.

Finally substitute the value of v2 back into your equation for v1 to find the value of v1.

I leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.

Answer 2

6.75x10^4