Conservation of momentum problem

• TheRedDevil18
In summary: The 4.9 is excluded because it is not a real number. It is an imaginary number. So the answer should have been 0.78s and the distance 7.8m, right ?
TheRedDevil18

Homework Statement

A 1200kg car traveling initially with a speed of 25m/s in an easterly direction crashes into the rear end of a 9000kg truck moving in the same direction at 20m/s. The velocity of the car right after the collision is 18m/s to the east.

3.4.1) What is the velocity of the truck after the collision
3.4.2) How much of mechanical energy is lost during the collision

p=mv
ek=1/2mv^2

The Attempt at a Solution

The velocity of the truck was worked out to be 21m/s. My problem is with the second question, this is an elastic collision right ?, so their should be no mechanical energy lost ?

TheRedDevil18 said:
The velocity of the truck was worked out to be 21m/s. My problem is with the second question, this is an elastic collision right ?, so their should be no mechanical energy lost ?

Ah, but presumably you used the given final velocity of the car to work out the final velocity of the truck using conservation of momentum, right? You didn't have to assume anything about the nature of the collision. Work out the kinetic energies before and after. What do you find?

Ok, this is what I get

ek before:
ek= 1/2*1200*25^ + 1/2*9000*20^2
= 2175000 J

ek after:
ek = 1/2*1200*18^2 + 1/2*9000*21^2
= 2178900 J

They are not the same, what is wrong ?

First, you should use more digits for intermediate values, particularly if they are going to be squared later on -- precision errors will be greatly multiplied by rounding or truncating too soon; The speed of the truck after the collision is not exactly 21 m/s. Use several more digits for the speed, and only round results for presentation (submission).

Second, they never said it was a perfectly elastic collision. (Nor, for that matter, did they say that it was a perfectly inelastic collision). So you can't say beforehand whether or not KE will be conserved.

Ok, thanks for the help. I also have another problem which I completely don't know where to start, here it is:

A stunt woman sitting on a tree branch wants to drop vertically onto a horse galloping under the tree. The horse gallops at a constant speed of 10m/s and the woman is initially 3m above the level of the saddle.

2.5.1) Calculate the horizontal distance between the saddle and the tree branch when the woman makes her move.

2.5.2) How long is she in the air ?

Next time post a separate thread for each problem and use the homework help template each time.

How much time is it going to take for the woman to fall 3 m? (Personally I wouldn't want to free fall 3 m if the landing was going to be on my behind).

How much horizontal distance does the horse travel in the amount of time you calculated above?

Therefore, how far ahead of the branch does the saddle have to be when she let's go?

Ok this is what I got:
I used the equation to find the time:
delta y = vi*t+1/2*a*t^2 and got my answer to be 0.35s.

I then used the equation d=st to get the horizontal distance of 3.5m.

Hmm, your equation looks correct but your answer for the free fall time is wrong. Can you post your work? What values did you use for 'a' and delta y?

For 'a' I used 9.8m/s and for delta y I used 3m.

You used the right values but got the wrong answer. So the problem is with your arithmetic, not with the physics. Unfortunately, you didn't post your calculation steps like I requested, so it's kind of hard for me to see what you did wrong or to correct it...

3 m = vit + (1/2)(9.81 m/s2)t2

What is vi, and how would you solve this equation for t?

vi is 0, am I right ?, anyway this is how I solved it:

3m = 0*t + (1/2)(9.8)t^2
3m = 4.9t^2
sqrt 3 = 4.9t
1.732 = 4.9t
t = 1.732/4.9
= 0.35s

I don't see anything wrong ?

TheRedDevil18 said:
vi is 0, am I right ?, anyway this is how I solved it:

3m = 0*t + (1/2)(9.8)t^2
3m = 4.9t^2
sqrt 3 = 4.9t
1.732 = 4.9t
t = 1.732/4.9
= 0.35s

I don't see anything wrong ?

Why is the 4.9 excluded from the square-rooting operation?

Sorry my silly mistake, so the answer should have been 0.78s and the distance 7.8m, right ?

What is the conservation of momentum principle?

The conservation of momentum principle is a fundamental law in physics that states that the total momentum of a system remains constant unless acted upon by an external force. This means that in a closed system, the total momentum before and after an interaction will be the same.

How is the conservation of momentum principle applied in real-life situations?

The conservation of momentum principle is applied in a wide range of real-life situations, such as collisions between objects, rocket propulsion, and even the movement of planets in our solar system. It also helps explain the motion of fluids and particles, such as in the flow of air and water.

What is an example of a conservation of momentum problem?

An example of a conservation of momentum problem would be a collision between two objects of different masses and velocities. By applying the principle, you can calculate the final velocities of the objects after the collision.

What are the key equations used in solving conservation of momentum problems?

The key equations used in solving conservation of momentum problems are the law of conservation of momentum, which states that the total momentum before and after an interaction is equal, and the momentum equation (p=mv), which relates an object's mass and velocity to its momentum. Other equations, such as the impulse-momentum theorem, can also be used in more complex problems.

What are some common misconceptions about the conservation of momentum principle?

One common misconception is that momentum is the same as velocity. While they are related, momentum also takes into account an object's mass. Another misconception is that conservation of momentum only applies to objects moving in a straight line. In reality, it applies to all types of motion, including rotational motion. Some also mistakenly believe that the conservation of momentum principle can be violated, when in fact it is a fundamental law of nature.

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