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Abbey and Mia are in the basement playing pool. On Abbey's recent shot, the cue ball was moving east at 91 cm/s when it struck the slower 8-ball moving in the same direction at 18 cm/s. The 8-ball immediately speeds up to 62 cm/s. Determine the post-collision velocity of the cue ball. (Assume all balls have the same mass.)

Sagot :

Jacob193idn

Answer:

[tex]47\; {\rm cm\cdot s^{-1}}[/tex] to the east.

Explanation:

When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].

Let [tex]m[/tex] denote the mass of one ball.

Before the collision:

  • Momentum of the cue ball: [tex]p(\text{cue, before}) = m\, v(\text{cue, before})[/tex].
  • Momentum of the numbered ball: [tex]p(\text{number, before}) = m\, v(\text{number, before})[/tex].

Hence, the total momentum before the collision was:

[tex]\begin{aligned} & p(\text{cue, before}) + p(\text{number, before}) \\ &= m\, v(\text{cue, before}) + m\, v(\text{number, before})\end{aligned}[/tex].

Likewise, the total momentum right after the collision would be:

[tex]\begin{aligned} & p(\text{cue, after}) + p(\text{number, after}) \\ &= m\, v(\text{cue,after}) + m\, v(\text{number, after})\end{aligned}[/tex].

Momentum is supposed to be conserved during the collision. In other words, total momentum should be the same immediately before and after the collision. Hence:

[tex]\begin{aligned} & p(\text{cue, before}) + p(\text{number, before}) \\ &= p(\text{cue, after}) + p(\text{number, after})\end{aligned}[/tex].

[tex]\begin{aligned} & m\, v(\text{cue, before}) + m\, v(\text{number, before}) \\ &= m\, v(\text{cue, after}) + m\, v(\text{number, after})\end{aligned}[/tex].

Assume that positive velocities denote motion to the east.

In this question, [tex]v(\text{cue, before})[/tex], [tex]v(\text{number, before})[/tex], and [tex]v(\text{number, after})[/tex] are all given. Rearrange this equation to find [tex]v(\text{cue, after})[/tex]:

[tex]\begin{aligned}& m \, v(\text{cue, after}) \\ &= m\, v(\text{cue, before}) \\ &\quad\quad + m\, v(\text{number, before}) \\ &\quad\quad - m\, v(\text{number, after}) \end{aligned}[/tex].

[tex]\begin{aligned}& v(\text{cue, after}) \\ &= v(\text{cue, before}) \\ &\quad\quad + v(\text{number, before}) \\ &\quad\quad - v(\text{number, after}) \\ &= 91\; {\rm cm \cdot s^{-1}} + 18\; {\rm cm\cdot s^{-1}} - 62\; {\rm cm\cdot s^{-1}} \\ &= 47\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Since the result is greater than zero, the direction of motion of the cue ball after the collision will also be to the east.

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