Velocity after a totally inelastic collision

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Homework Help Overview

The problem involves a totally inelastic collision between a car and a bug, where the car has a mass of 1000 kg and is traveling at 25 m/s, while the bug has a mass of 9.0 g and is moving at -1.5 m/s. The objective is to determine the change in velocity of the car after the collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and the correct setup of the equation for final velocity. There are questions about the clarity of the problem statement and the correctness of the initial attempts at a solution.

Discussion Status

Some participants have confirmed the question being asked and noted that previous answers were incorrect. Guidance has been offered regarding the equation for final velocity, and there is an ongoing exploration of how to manipulate the equation without evaluating it fully.

Contextual Notes

Participants are navigating the nuances of the problem, including the small mass of the bug relative to the car and the implications for the change in velocity. There is a focus on ensuring the correct application of the momentum conservation principle.

Manh
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Homework Statement


You are driving your 1000-kg car at a velocity of(25 m/s )ι^ when a 9.0-g bug splatters on your windshield. Before the collision, the bug was traveling at a velocity of (-1.5 m/s )ι^.
What is the change in velocity of the car due to its encounter with the bug?

Homework Equations


pi = pf
m1v1 + m2v2 = (m1 + m2)v

The Attempt at a Solution


p1 + p2 = (m1 + m2)v
(2.5 x 10^4) + (-1.35 x 10^-2) = (1000 + 0.009)v
v = 25 m/s
 
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So is there a question here?
 
Chestermiller said:
So is there a question here?
Yes. The question is "What is the change in velocity of the car due to its encounter with the bug?". I also came up with an answer but it was incorrect.
 
Manh said:
Yes. The question is "What is the change in velocity of the car due to its encounter with the bug?". I also came up with an answer but it was incorrect.
You got the correct equation for the final velocity, and I am just going to re-write it out for you as follows:

$$v=\frac{(2.5 \times 10^4-1.35 \times 10^{-2})}{1000+0.009}$$

If you subtract the original velocity of the car, you get the change in velocity Δv:

$$Δv=\frac{(2.5 \times 10^4-1.35 \times 10^{-2})}{1000+0.009}-25$$

Now, what I would like you to do is to reduce the relationship to a common denominator, without first evaluating the first term and without combining terms in the numerator. What do you get?

Chet
 

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