Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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 :

jacob193

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.

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.