Question

1. A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0e from its original direction. What is the speed of the initially stationary billiard ball after the collision?
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Answer #1

Nomenclature is like this

u1xi = initial speed of moving ball in x-direction

u2xi = initial speed of stationary ball in x-direction

u1yi = initial speed of moving ball in y-direction

u2yi = initial speed of stationary ball in y-direction

V1 = final speed of initially moving ball

V2 = final speed of initially stationary ball

In a perfectly elastic collision

Suppose final velocity of stationary billiard ball is v2 and it's making angle \theta with horizontal axis
Using momentum conservation in x direction,
Pi = Pf
m1*u1xi + m2*u2xi = m1*V1x + m2*V2x
given that m1 = m & m2 = m

u1xi = 3 m/sec & u2xi = 0 (since stationary)

V1x = V1*cos 30 deg & V2 = V2*cos \theta
m*u1xi + m*u2xi = m*V1*cos(30 deg) + m*V2*cos \theta
u1xi + u2xi = V1*cos(30 deg) + V2*cos \theta
3 - 0 = V1*cos(30 deg) + V2*cos \theta
V1*cos(30 deg) + V2*cos \theta = 3

V2*cos \theta = 3 - V1*cos 30 deg

Using momentum conservation in y direction,
Pi = Pf
m1*u1yi + m2*u2yi = m1*V1y + m2*V2y
given that m1 = m & m2 = m

u1yi = 0 m/sec & u2yi = 0 m/sec

V1y = V1*sin 30 deg & V2y = -V2*sin \theta
m*u1i + m*u2i = m*V1*sin(30 deg) - m*V2*sin \theta
u1i + u2i = V1*sin(30 deg) - V2*sin \theta
0 - 0 = V1*sin(30 deg) - V2*sin \theta

V1*sin(30 deg) = V2*sin \theta

V2*sin \theta = (1/2)*V1

Now square and add both equation

V2^2*(sin2\theta + cos2\theta) = (1/4)*V1^2 + 9 + (3/4)*V1^2 - (3*sqrt 3)*V1

V2^2*(1) = (1/4)*V1^2 + 9 + (3/4)*V1^2 - (3*sqrt 3)*V1

V2^2 = (1/4)*V1^2 + 9 + (3/4)*V1^2 - (3*sqrt 3)*V1

V2^2 = V1^2 + 9 - (3*sqrt 3)*V1

Now Using energy conservation in elastic collision

KEi = KEf

0.5*m*u1^2 + 0.5*m*u2^2 = 0.5*m*V1^2 +0.5*m*V2^2

u1^2 + u2^2 = V1^2 + V2^2

V1^2 + V2^2 = 3^2 + 0

V1^2 + V2^2 = 9

V2^2 = 9 - V1^2

Using above expression

9 - V1^2 = V1^2 + 9 - (3*sqrt 3)*V1

2*V1^2 - (3*sqrt 3)*V1 = 0

V1 = (3*sqrt 3)/2

V1 = 2.598 m/sec = 2.6 m/sec

So,

V2 = sqrt (9 - V1^2)

V2 = sqrt (9 - 2.598^2)

V2 = 1.50 m/sec = final speed of stationary billiard ball

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