Velocity after a totally inelastic collision

In summary: I cannot provide a summary for this conversation as it is not related to the initial content. Can you provide a summary for the conversation about the change in velocity of the car due to its encounter with the bug?In summary, we discussed the change in velocity of a car due to its encounter with a bug on the windshield. The correct equation for the final velocity was provided, and then we derived the equation for the change in velocity by subtracting the original velocity of the car. We were then asked to reduce the relationship to a common denominator without evaluating the first term or combining terms in the numerator.
  • #1
Manh
62
0

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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  • #2
So is there a question here?
 
  • #3
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.
 
  • #4
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
 

Related to Velocity after a totally inelastic collision

1. What is a totally inelastic collision?

A totally inelastic collision is a type of collision where two objects stick together after colliding. This means that they move together as one object after the collision.

2. How is the velocity calculated after a totally inelastic collision?

The velocity after a totally inelastic collision can be calculated using the equation v = (m1v1 + m2v2)/(m1 + m2), where v is the final velocity, m is the mass of the object, and v is the initial velocity.

3. What is the conservation of momentum in a totally inelastic collision?

The conservation of momentum states that the total momentum of a closed system remains constant before and after a collision. In a totally inelastic collision, the total momentum of the two objects before the collision is equal to the total momentum of the combined object after the collision.

4. How does the mass of the objects affect the velocity after a totally inelastic collision?

The mass of the objects has a direct impact on the velocity after a totally inelastic collision. The heavier object will have a larger initial velocity, and after the collision, the combined object will have a lower velocity compared to the lighter object.

5. Can the velocity after a totally inelastic collision be greater than the initial velocity of either object?

No, the velocity after a totally inelastic collision can never be greater than the initial velocity of either object. This is because some kinetic energy is lost during the collision, resulting in a decrease in velocity.

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