Question

One billiard ball is shot east at 1.9m/s . A second, identical billiard ball is shot west at 0.90m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90? and sending it north at 1.31m/s .

1.)What is the speed of the first ball after the collision?

2.)What is the direction of the first ball after the collision? Give the direction as an angle south of east.

Answer 1

According to law of conservation of momentum, the total momentum of the system along x-axis is

m1 v1 + m2 v2 = m1 u1 cos (theta )           ..... (1)

net momentum ofthe system along y -axis is

0 = m1 u1 sin (theta) + m2 u2

   - m2 u2 = m1 u1 sin(theta) .....(2)

divide equation (2) by equation (1)

            tan (theta ) = - m2 u2 / ( m1 v1 + m2 v2 )

since m1 = m2 = m

             tan (theta ) =   -u2 / ( v1 + v2 ) = - 1.31 / ( 1.9 + 0.9 )

                     theta =- 25 degrees

velocity of first billiard ball is

         u1 = m1 v1 + m2 v2 / m1 u1 cos (theta )

              = ( 1.9 m/s + 0.9 m/s) / (1.31 m/s ) cos (-25 )

              =2.35 m/s

direction first ball is -25 degrees or 335 degrees from positive x-axis.