Question

A particle moving at 20 m/s collides elastically with an identical particle at rest. the incoming particle moves off at 30 to the original direction. What is the final speed of each particle?

Answer 1

here,

the initial speed of 1 , u1 = 20 m/s

theta1 = 30 degree

let their final speeds be v1 and v2 and the angle for velocity of seccond particle be theta2

along the original line of motion

using conservation of momentum

m * u1 = m * v1 * cos(theta1) + m * v2 * cos(theta2)

20 = v1 * cos(30) + v2 * cos(theta2) .....(1)

and

along the line perpendicular to the original line of motion

using conservation of momentum

0 = m * v1 * sin(theta1) - m * v2 * sin(theta2)

0 = v1 * sin(30) - v2 * sin(theta2) .....(2)

and

using conservation of kinetic energy

0.5 * m * u1^2 = 0.5 * m * v1^2 + 0.5 * m * v2^2

20^2 = v1^2 + v2^2 ....(3)

from (1) , (2) and (3)

v1 = 3.09 m/s

v2 = 19.76 m/s

theta = 355.5 degree from the original line of motion

the final speed of initial moving particle is 3.09 m/s

the final speed of second particle is 19.76 m/s and 355.5 degree from the original line of motion