Question

Two identical pucks are on an air table. Puck A has an initial velocity of 2.5 m/s in the positive x-direction. Puck B is at rest. Puck A collides elastically with puck B and A moves off at 1.4 m/s at an angle of +60

Answer 1

here ,

we will use conservation of momentum ,

using conservation of momentum in x - axis ,

vxi*m = vxf * m + m*1.4 * cos(60)

2.5 = vxf + 0.7

vxf = 1.8 m/s

Now , balacing momentum in y - axis

vyf * m + m * 1.4 *sin(60) = 0

vyf = -1.21 m/s

Now ,

Vf = 1.8 i - 1.21 j

magnitude of final velocity = sqrt(1.8^2 + 1.21^2)

magnitude of final velocity = 2.17 m/s

theta = arctan(-1.21/1.8)

theta = - 33.9 degree

magnitude of final velocity is 2.17 m/s at theta = - 33.9 degree