Velocity after a totally inelastic collision

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SUMMARY

The discussion centers on calculating the change in velocity of a 1000-kg car after a totally inelastic collision with a 9.0-g bug. The initial velocities are 25 m/s for the car and -1.5 m/s for the bug. The conservation of momentum equation is applied, leading to the formula for final velocity: v = (m1v1 + m2v2) / (m1 + m2). The change in velocity (Δv) is derived by subtracting the initial velocity from the final velocity, resulting in Δv = [(2.5 x 10^4 - 1.35 x 10^-2) / (1000 + 0.009)] - 25.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with inelastic collisions
  • Basic algebra for manipulating equations
  • Knowledge of unit conversions (grams to kilograms)
NEXT STEPS
  • Study the principles of inelastic collisions in physics
  • Learn about momentum conservation in multi-object systems
  • Practice solving problems involving momentum and velocity changes
  • Explore the effects of mass differences in collision scenarios
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Physics students, educators, and anyone interested in understanding the dynamics of collisions and momentum conservation in real-world scenarios.

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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