1. Dec 21, 2009

### FAJAS

1. The problem statement, all variables and given/known data

Q1.A person of mass 80kg is travelling in a car with a velocity of 25m.s the car is involved in a head on collision and brought rapidly to a halt. Neglecting frictional forces effects. Calculate (a) a person momentum just prior to impact (b) the resulting force if the impact time is 0.005s

Q2.A 70kg person falls from a height of 1m if they land 'stiff legged' the body will come to rest in 0.005s what force will the person's leg experience?

Q3.A 5kg mass falls through 5m and falls on pedestrians head coming to a halt within 0.2s of the initial impact. calculate the force of impact on the pedestrians skull.

2. Relevant equations

force(change in time) = mass(change in velocity)

3. The attempt at a solution

I just have the answers to these questions I need some explanation !!

2. Dec 21, 2009

### diazona

3. Dec 21, 2009

### FAJAS

*For the first question i used the impulse equation and momentum equations
(a) p=mv
80x25 = 2000kg/m.s
(b) FT=m(v-u)
=80(0-25)/0.005
= -400000N

*For the second question i used the velocity/displacement equation to find out the velocity then i used the impulse equation just like the one above to solve for the Force.
The answer i got was 2800000N

*For the third question i did the same thing as i did with the second question. My answer was 5N.

These are my answers hope you could help correct me if i'm wrong :)

4. Dec 21, 2009

### arunma

$$F\Delta t = mv$$

You also know that momentum is,

$$p = mv$$

Solving problems one and two with these equations is just a matter of plug 'n chug.

For the third problem, here's a hint. Use the kinematic equation of motion that you likely learned earlier in the semester,

$$v^2 = v_0 ^2 + 2a(x-x_0)$$

That'll give you the speed at impact, from which you can use the first equation I mentioned to find the force of impact.

5. Dec 21, 2009

### FAJAS

so the answer is 250N ?

6. Dec 22, 2009

### arunma

Oh I'm sorry, I didn't see your previous post.

You did the first problem correctly (there's no minus sign required on the force, but I doubt any grader would take away points for that). You seem to have gotten something wrong on the second equation. I would take another look at your "velocity displacement equation." In the last equation in my previous post, all you need to do is solve for the velocity (which requires only a single step). And I assume that the 250 N is your answer to the last problem. If so, then you're correct.

$$F = \dfrac{m\sqrt{2g(x-x_0)}}{\Delta t}$$

and then you just happened to plug in the wrong numbers, you'll still get most of the points. Heck, if I were grading, I'd give full credit to someone who had the symbolic answer right but screwed up the numbers.

Also it'll make it easier for us to figure out how to help you on the second problem.

7. Dec 22, 2009

### FAJAS

i used this formula(the one you had in the previous post) now and got 62609.90N
Am i on the right track or still mssing up something?

(i always write down all my steps but im not used to typing them out )

I have more questions if you could help?!

8. Dec 22, 2009

### pgardn

Are you using g = 10 m/s/s? Just so I can check...
You are on the right track.
Not to be mean, but I think this belongs in the "regular" Physics section.
I dont think this is upper level undergraduate Physics.

For #3, I see a mass right before it hits the persons head moving down at a speed of ____ coming to rest for a final velocity of zero. The force of the persons head accelerated the mass up over a period of 0.2 seconds and brought the 5 kg mass to rest. So mass X change in V all divided by 0.2 seconds, yes?

For the blank above, what is the velocity of any object if it has fallen 5 m (assume starting from rest)?

Last edited: Dec 22, 2009
9. Dec 22, 2009

### FAJAS

i got 62609.90N for the second question and 250N for the third. Yes i im using 10m/s/s as the acceleration.

10. Dec 22, 2009

### pgardn

Since they are the same lets check #3.

What did you get for the change in velocity? I got vf = 0 m/s and just before the object hits the persons head and comes to rest, I got the mass going -10 m/s.
Ok gotta go, I got 250N. Do number 2 the same way. And it would be negative if is the force of the mass on the persons head. If it was the force of the head on the object it would be positive. Taking up to be +...

Last edited: Dec 22, 2009
11. Dec 22, 2009

### FAJAS

i got 250N too but the velocity was root 100 so it could be either negative or positive 10 but i just used 10 not the negative 10. I never thought of that about whats falling and stuff like that i just know that the acceleration would be positive if we go upwards and negative if we go downwards but thats for the effective weight thing. I dont know this is just confusing.
Regarding the second question i got the same answer as i did before 62609.9N. i did exactly the same thing as i did for the third question. ( the number seems so large i think its WRONG ! :s )

12. Dec 22, 2009

### pgardn

Well you got a larger mass coming to a halt in a much shorter period of time. So it requires a large force to do this.
I personally like to keep in mind the actual physical process that is occurring and remembering that most of this mechanics falls back on Newtons laws. That way when I read a question, its not just plugging numbers into an equation. It makes physical and mathematical sense.

In collisions the force between two objects must involve the same amount of force for the same amount of time thus the same impulse is exerted on each of the two objects, just in different directions (third law). And then it makes sense that both objects must undergo exactly the same change in momentum (F*t), even though the velocities involved and masses may be different. Furthermore if we just consider the just the force due to contact during the collision, it makes perfect sense that both objects should be accelerated. And if they are accelerated a bunch(big change in velocity) over a short period of time, it would make the force large, especially if the object was massive. F=ma second law. It all fits together quite nicely.

I must see the physical process and then the ideas and math that go with Newton's Laws fits. So problems are much easier to solve. You can turn them any way you want to and they still make sense.

By the way if you set F*t = (-)F*t for these collisions, the conservation of momentum appears to make sense. Ahh but I digress. I like getting more than the right answer out of this stuff.

13. Dec 22, 2009

### FAJAS

so is the answer to the second question correct ? :P