Velocity after a totally inelastic collision

  • Thread starter Manh
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  • #1
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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
 

Answers and Replies

  • #3
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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.
 
  • #4
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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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