Question

A billiard ball is shot east at 2.80 m/s. A second, identical billiard ball is shot west at 1.20 m/s. The balls has a glancing collision, not a head-on-collision, deflecting the second ball by 90° and sending it north at 1.60 m/s. What is the angle that the velocity of the first ball makes after the collision, with respect to the east direction?

Answer 1

m1 = m kg                      m2 = m kg

before collision

speeds

u1x = 2.8 m/s                        u2x = -1.2 m/s

u1y = 0                                u2y = 0

after collision

v1x = ?                     v2x = 0

v1y = ?                     v2y = 1.6 m/s

from momentum conservation

along y momentum is conserved

momentum beifre collision = momentum after collision

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

0 = m*v1y + m*1.6

v1y = - v2y = -1.6 m/s

along x axis momentum is conserved

momentum before collision = momentum after collision

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m*2.8 - m*1.2 = m*v1x + 0

1.6 = v1x

v1x = 1.6 m/s

v1 = sqrt(v1x^2+v1y^2)

v1 = 2.26 m/s

direction = tan^-(v1y/v1x) = 270 degrees