Question

One billiard ball with a mass of 0.50 kg is shot east at 2.5 m/s. A second billiard ball with a mass of 0.25 kg is shot west at 2.0 m/s. The balls have a glancing collision, not a head-on collision, deflecting the first ball by 90° and sending it north at 1.0 m/s.

What are the speed of the second ball after the collision?

What is the direction of the second ball after the collision? Express your answer as an angle in degrees cw from East. Use deg for degrees.

Answer 1

here,

mass of ball 1, m1 = 0.5 kg

mass of ball 2 , m2 = 0.25 kg

initial speed of 1 , u1 = 2.5 i m/s

initial speed of 2 , u2 = - 2 i m/s

final speed of ball 1 , v1 = 1 j m/s

let the final speed of seccond be v2

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

0.5 * 2.5 i - 0.25 * 2 i = 0.5 * 1 * j + 0.25 * v2

v2 = 3 i - 2 j m/s

the magnitude of final velocity , |v2| = sqrt(3^2 + 2^2) = 3.61 m/s

the direction of final velocity , theta = arctan(2/3) = 33.7 degree south of east