Question

A cue ball traveling at 4.0 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30° with its original direction of travel.

(a) Find the angle between the velocity vectors of the two balls after the collision. °

(b) Find the speed of each ball after the collision. cue ball m/s target ball m/s

Answer 1

a)

m = mass of each ball

vcxo = initial velocity of cue ball along x-direction = 4 m/s

vtxo = initial velocity of target ball along x-direction = 0 m/s

vcxf = final velocity of cue ball along x-direction = 4 Cos30 m/s

vtxf = final velocity of target ball along x-direction = vt Cos

Using conservation of momentum

m vcxo + m vtxo = m vcxf + m vtxf

4 + 0 = vc Cos30 + vt Cos

vt Cos = 4 - vc Cos30   Eq-1

vcyo = initial velocity of cue ball along y-direction = 0 m/s

vtyo = initial velocity of target ball along y-direction = 0 m/s

vcyf = final velocity of cue ball along y-direction = vc Sin30 m/s

vtyf = final velocity of target ball along y-direction = - vt Sin

Using conservation of momentum

m vcyo + m vtyo = m vcyf + m vtyf

0 + 0 = vc Sin30 - vt Sin

vt Sin = vc Sin30 Eq-2

Using conservation of kinetic energy

m v2co + m v2to = m v2cf + m v2tf

v2co + v2to = v2cf + v2tf

42 + 02 = vc2 + vt2

vc2 + vt2 = 16 Eq-3

Squaring and adding Eq-1 and Eq-2

v2t Cos2 + v2t Sin2 = (4 - vc Cos30)2 + v2c Sin230

v2t = (4 - vc Cos30)2 + v2c Sin230

Using Eq-3

16 - vc2 = (4 - vc Cos30)2 + v2c Sin230

vc = 3.5 m/s

Using Eq-3

vc2 + vt2 = 16

3.52 + vt2 = 16

vt = 1.94 m/s

using Eq-2

vt Sin = vc Sin30

1.94 Sin = 3.5 Sin30

Sin = 0.9021

= Sin-1(0.9021)

= 64.44 deg

Angle between the two balls is

64.44 + 30

94.44 deg

b)

vc = 3.5 m/s

vt = 1.94 m/s