Question

A novice pool player is faced with the corner pocket shot shown in Fig. 7–49. Relative dimensions are also shown. Should the player worry that this might be a “scratch shot,” in which the cue ball will also fall into a pocket? Give details. Assume equal-mass balls and an elastic collision. Ignore spin.

Short answer

The cue ball will also fall into the same pocket.

Step-by-step solution

Identification of the given data 

The width of the pool table is 4 units.

The vertical distance of the red ball is\(\sqrt 3 \)units.

The horizontal distance of the red ball is 1 unit.

Statement of the principle of conservation of linear momentum

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after the collision remains the same if no external force acts on the system.It is given as:

\(\begin{aligned}{p_{{\rm{before}}}} &= {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} &= {\left( {mv} \right)_{{\rm{after}}}}\end{aligned}\)

Statement of the principle of conservation of energy

According to the principal of conservation of energy, the total energy of the system before the collision is equal to the total energy of the system after the collision.

It is given as:

\({\left( {KE + PE} \right)_{{\rm{before}}}} = {\left( {KE + PE} \right)_{{\rm{after}}}}\)

Applying the conservation of linear momentum

Suppose, the red ball makes an angle of and the cue ball makes an angle of with the vertical. Let the final speed of the red ball be , and that of the cue ball be . Since the collision is elastic, apply the conservation of linear momentum.

Along the x-axis:

\(\begin{aligned}{c}{\left( {{p_{\rm{x}}}} \right)_{{\rm{before}}}} &= {\left( {{p_{\rm{x}}}} \right)_{{\rm{after}}}}\\mv &= \left( {\left( {m{v_1}^\prime \cos \theta } \right) + \left( {m{v_2}^\prime \cos \alpha } \right)} \right)\\v &= \left( {\left( {{v_1}^\prime \cos \theta } \right) + \left( {{v_2}^\prime \cos \alpha } \right)} \right)\\{v^2} &= \left( {\left( {{v_1}{{^\prime }^2}{{\cos }^2}\theta } \right) + \left( {2{v_1}^\prime {v_2}^\prime \cos \theta \cos \alpha } \right) + \left( {{v_2}{{^\prime }^2}{{\cos }^2}\alpha } \right)} \right)\end{aligned}\) … (i)

Along the y-axis:

\(\begin{aligned}{c}{\left( {{p_{\rm{y}}}} \right)_{{\rm{before}}}} &= {\left( {{p_{\rm{y}}}} \right)_{{\rm{after}}}}\\0 &= \left( {m{v_1}^\prime \sin \theta - m{v_2}^\prime \sin \alpha } \right)\\0 &= \left( {{v_1}^\prime \sin \theta - {v_2}^\prime \sin \alpha } \right)\\0 &= \left( {{v_1}{{^\prime }^2}{{\sin }^2}\theta - 2{v_1}^\prime {v_2}^\prime \sin \theta \sin \alpha + {v_2}{{^\prime }^2}{{\sin }^2}\alpha } \right)\end{aligned}\) … (ii)

Adding equations (i) and (ii), you get:

\({v^2} = \left( {{v_1}{{^\prime }^2} + {v_2}{{^\prime }^2} + 2{v_1}^\prime {v_2}^\prime \cos \left( {\theta + \alpha } \right)} \right)\) … (iii)

Applying the conservation of kinetic energy

As the collision is elastic, the kinetic energy of the system will be conserved and the objects will fly apart separately after collision. Therefore, the white (cue) ball will not travel the same direction as the red ball and fall into the same pocket.

Applying the law of conservation of energy, you get:

Along the x-direction:

\(\begin{aligned}{\left( {K{E_x}} \right)_{before}} &= {\left( {K{E_x}} \right)_{after}}\\ \frac{1}{2}m{v^2} &= \left( {\left( {\frac{1}{2}m{v_1}{{^\prime }^2}{{\cos }^2}\theta } \right) + \left( {\frac{1}{2}m{v_2}{{^\prime }^2}{{\cos }^2}\alpha } \right)} \right)\\ m{v^2} &= \left( {\left( {m{v_1}{{^\prime }^2}{{\cos }^2}\theta } \right) + \left( {m{v_2}{{^\prime }^2}{{\cos }^2}\alpha } \right)} \right)\\ {v^2} &= \left( {\left( {{v_1}{{^\prime }^2}{{\cos }^2}\theta } \right) + \left( {{v_2}{{^\prime }^2}{{\cos }^2}\alpha } \right)} \right) \end{aligned}\)

… (iv)

Along the y-direction:

\(\begin{aligned}{\left( {K{E_y}} \right)_{before}} &= {\left( {K{E_y}} \right)_{after}}\\ 0 &= \left( {\left( {\frac{1}{2}m{v_1}{{^\prime }^2}{{\sin }^2}\theta } \right) + \left( {\frac{1}{2}m{v_2}{{^\prime }^2}{{\sin }^2}\alpha } \right)} \right)\\ 0 &= \left( {\left( {m{v_1}{{^\prime }^2}{{\sin }^2}\theta } \right) + \left( {m{v_2}{{^\prime }^2}{{\sin }^2}\alpha } \right)} \right)\\ 0 &= \left( {\left( {{v_1}{{^\prime }^2}{{\sin }^2}\theta } \right) + \left( {{v_2}{{^\prime }^2}{{\sin }^2}\alpha } \right)} \right) \end{aligned}\)

… (v)

Adding equations (4) and (5), you get:

\(\begin{aligned}{c}{v^2} &= \left( {{v_1}{{^\prime }^2}{{\cos }^2}\theta } \right) + \left( {{v_2}{{^\prime }^2}{{\cos }^2}\alpha } \right) + \left( {{v_1}{{^\prime }^2}{{\sin }^2}\theta } \right) + \left( {{v_2}{{^\prime }^2}{{\sin }^2}\alpha } \right)\\{v^2} &= \left( {{v_1}{{^\prime }^2}} \right) + \left( {{v_2}{{^\prime }^2}} \right)\end{aligned}\) … (vi)

Compare equations (iii) and (vi).

\(\begin{aligned}{c}\left( {{v_1}{{^\prime }^2} + {v_2}{{^\prime }^2}} \right) &= \left( {{v_1}{{^\prime }^2}} \right) + \left( {{v_2}{{^\prime }^2}} \right) + \left( {2{v_1}^\prime {v_2}^\prime \cos \left( {\theta + \alpha } \right)} \right)\\2{v_1}^\prime {v_2}^\prime \cos \left( {\theta + \alpha } \right) = 0\\\cos \left( {\theta + \alpha } \right) &= 0\\\theta + \alpha &= {90^{\rm{o}}}\end{aligned}\) … (vii)

From the given diagram, you get:

\(\begin{aligned}{c}\tan \theta = \frac{1}{{\sqrt 3 }}\\\theta = {30^{\rm{o}}}\end{aligned}\)

Substitute the values in equation (vii).

\(\begin{aligned}{c}{30^{\rm{o}}} + \alpha = {90^{\rm{o}}}\\\alpha = {60^{\rm{o}}}\end{aligned}\)

Thus, if the angles made by the balls are \({30^{\rm{o}}}\) and \({60^{\rm{o}}}\), the cue ball will enter the right pocket.