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Competition billiard balls and tables need to adhere to strict standards (see the Billiard Congress of America for standards in the United States). Specifically, billiard balls must weigh between 5.5 and 6 oz, and they must be 2.25\pm 0.005 in. in diameter.

Using the theory presented in this section, establish whether or not it is possible to have a moving ball A hit a stationary ball B so that A stops right after the impact, if A and B have the same diameter but not the same weight (since it appears possible to have a weight difference of up to 0.5 oz while staying within regulations). Assume that the COR e = 1.

Step-by-step

If the pre-impact velocity of A had a non-zero component of velocity perpendicular to the LOI, then this component of velocity would be conserved through the impact. This consideration implies that for A to stop after impact, at the very least, its preimpact velocity must be entirely parallel to the LOI. Working under this assumption and using the component system shown, the preimpact velocities of A and B are

{ \overrightarrow { \upsilon } }_{ A }^{ – }={ \upsilon }_{ A }^{ – }\widehat { \jmath }  and\quad { \overrightarrow { \upsilon } }_{ B }^{ – }=\overrightarrow { 0 }

where { \upsilon }_{ A }^{ – } is the preimpact speed of A, having assumed that A is initially moving in the positive y direction. Consequently, the conservation of linear momentum in the y direction reads

{ m }_{ A }{ \upsilon }_{ A }^{ – }={ m }_{ A }{ \upsilon }_{ Ay }^{ + }+{ m }_{ B }{ \upsilon }_{ By }     (1)

and the COR equation reads

{ \upsilon }_{ By }^{ + }-{ \upsilon }_{ Ay }^{ + }=e{ \upsilon }_{ A }^{ – }     (2)

Equations (1) and (2) are two equations in the two unknowns { \upsilon }_{ Ay } and { \upsilon }_{ By }, whose solution is

{ \upsilon }_{ Ay }^{ + }=\frac { ({ m }_{ A }-mBe){ \upsilon }_{ A }^{ – } }{ { m }_{ A }+{ m }_{ B } } \quad and\quad { \upsilon }_{ By }^{ + }=\frac { { m }_{ A }{ \upsilon }_{ A }^{ – }(e+1) }{ { m }_{ A }+{ m }_{ B } }     (3)

To check whether or not it is possible for A to stop, we set { \upsilon }_{ Ay }^{ + }=0 in Eq. (3), then we have

\frac { ({ m }_{ A }-mBe){ \upsilon }_{ A }^{ – } }{ { m }_{ A }+{ m }_{ B } } =0\Longrightarrow { m }_{ A }={ m }_{ B }e     (4)

Recalling that the COR e = 1, we have that { \upsilon }_{ Ay } can only be equal zero if the masses are identical.

Given the assumptions, it is not possible to have a moving ball A hit a stationary ball B so that A stops right after the impact.

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