In a game of billiards ball A is given an initial velocity v

In a game of billiards, ball A is given an initial velocity v_0 along the longitudinal axis of the table. It hits ball B and then ball C. which are both at rest. Balls A and C are observed to hit the sides of the table squarely at A\' and C. respectively, and ball B is observed to hit the side obliquely at B. Knowing that v_0 = 5 m/s, v_A = 1.92 m/s and a = 1.65 m, determine the velocities v_B and v_C of balls B and C and the Point G where ball C hits the side of the table. Assume frictionless surfaces and perfectly elastic impacts (that is. conservation of energy). The velocity of ball B is (Round the final answers to two decimal places.) The velocity of ball C is (Round the final answers to two decimal places.) The Point C\' where ball C hits the side of the table is (Round the final answers to two decimal places.)

Solution

By energy conservation:

1/2*m_a*Vo^2 = 1/2*m_a*V_a^2 + 1/2*m_b*V_b^2 + 1/2*m_c*V_c^2

Assuming masses of all the balls to be equal, m_a = m_b = m_c = m

1/2*m*Vo^2 = 1/2*m*V_a^2 + 1/2*m*V_b^2 + 1/2*m*V_c^2

Vo^2 = V_a^2 + V_b^2 + V_c^2

5^2 = 1.92^2 + V_b^2 + V_c^2

V_b^2 + V_c^2 = 21.3136.................eqn1

Let angle between V_b and Vo = theta

By conservation of momentum along the vertical axis,

0 = m*V_b*sin theta - m*V_a

0 = V_b*sin theta - V_a

0 = V_b*sin theta - 1.92

V_b*sin theta = 1.92.................eqn2

By conservation of momentum along the longitudinal axis,

m*Vo = m*V_b*cos theta + m*V_c

Vo = V_b*cos theta + V_c

5 = V_b*cos theta + V_c

V_b*cos theta = 5 - V_c....................eqn3

Squaring eqns 2 and 3 and adding them gives,

V_b^2 = 1.92^2 + (5 - V_c)^2

V_b^2 = V_c^2 - 10*V_c + 28.6864

From eqn 1, V_b^2 = 21.3136 - V_c^2

Therefore, 21.3136 - V_c^2 = V_c^2 - 10*V_c + 28.6864

2*V_c^2 - 10*V_c + 7.3728 = 0

Solving this quadratic eqn we get V_c = 4.1 m/s or 0.9 m/s

If we take V_c = 4.1 m/s we get

From eqn3, V_b*cos theta = 0.9 m/s

Dividing it from eqn2, we get Tan theta = 1.92 / 0.9 or theta = 64.88 deg

and V_b = 1.92 / sin64.88 = 2.12 m/s

If we take V_c = 0.9 m/s we get

From eqn3, V_b*cos theta = 4.1 m/s

Dividing it from eqn2, we get Tan theta = 1.92 / 4.1 or theta = 25.09 deg

and V_b = 1.92 / sin25.09 = 4.527 m/s

Additionally assuming that after colliding with B, the velocity of A is V and the angle between BC and longitudinal axis is phi.

The following eqns of motion also need to be satisfied:

Vb*cos theta + V*cos phi = 5
Vb*sin theta = V*sin phi = 1.92
V*cos phi = Vc

If Vc = 4.1 m/s, we get phi = 25.09 deg and V = 4.527 m/s

If Vc = 0.9 m/s, we get phi = 64.88 deg and V = 2.12 m/s

Taking initial position of B from left end be b we get,

Tan theta = 0.75 / (1.8 - b)

Tan phi = (c - 0.75) / (1.65 - b)

If Vc = 0.9 m/s which implies theta = 25.09 deg and phi = 64.88 deg, we get b = 0.198 m, c = 3.846 m

But c cannot be greater than the width of the table 1.5 m.

Thus Vc = 4.1 m/s which implies theta = 64.88 deg and phi = 25.09 deg, we get b = 1.44 m, c = 0.844 m

Hence,

We get,

V_b = 2.12 m/s, theta = 64.88 deg

V_c = 4.1 m/s,

c = 0.844 m