# Homework Help: How to find initial speed of something without time or acceleration

1. Apr 21, 2012

### physical..O.o

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

Hi..heres the problem:
A 4WD travelling East has collided with a stationary car, the 4WD's skid marks are 15m long until the point of collision. The 4WD is sent 35degs South for 15.8m and the car is sent 55degs North for 15m. The car has a mass of 1150kg and the 4WD's is 1600kg. .The coefficient of friction is 0.6.
Find the initial speed of the 4WD before braking in km/h.
this is ALL the info included...

2. Relevant equations

s=displacement, mu=0.6, v=final velocity, u=initial velocity, P=momentum, Pi=initial momentum (also just the momentum of 4WD before crash), Pf=final momentum, P'1=momentum of 4WD after collison, P'2=momentum of car after collision, g=gravity(9.8plz), a=acceleration t=time ( numbers after pronumerals signify which vehicle: 4WD=1 , Car=2...dashes between pronumeral and vehicle number mean after the crash eg: P'1)

u= sqr root ( 2mu*g*s)

P=mv

pythagorus= to add the final momentums to get the initial momentum

v^2=u^2 + 2as

s=ut+1/2 at^2 ....(1/2 = half)

v=u+at

3. The attempt at a solution

to find initial speeds after the crash:
u'1= sqr root:(2*0.6*9.8*15.8)
u'1= 13.63m/s

u'2= sqr root:(2*0.6*9.8*15*)
u'2= 13.28m/2

to find the initial momentum (initial momentum of 4WD)
Pi^2 = (m'1*u'1)^2 + (m'2*u'2)^2
Pi^2=(1600*13.63)^2 + (1150*13.28)^2
Pi= 26623.727 kgm/s
then to find v1= (26623.727)/(1600)
v1= 16.63m/s

from there on i dont know what to do.. I've been advised that the last answer (16.63m/s) was the final speed before the collison...is that right? if so, how can i find the initial speed before braking? im guessing i have to find acceleration, but i dont know how to without time, and visa versa... apparently the initial speed is in the 70's km/h...:S plz help

2. Apr 21, 2012

### physical..O.o

if anyone thinks this question cant be done please say so so i havnt got my hopes up...but it should be, others have got the answer, i just have no idea how to get it.. iv spent ages on it..:(

3. Apr 21, 2012

### physical..O.o

cmon guysss help me out here...its really urgent! D:

4. Apr 21, 2012

### ehild

You found the speed of both cars after the collision. Apply conservation of momentum for the process of collision: write it up for both components: x (East) and y (North-South)

The x component of the initial momentum of the 4WD is equal to the x component of its momentum after collision plus the x component of the momentum of the car.
The y components of the momenta are equal and opposite after collision.
You get the components of momenta from their magnitude and the angle they enclose with the Eastward direction after collision.

ehild

5. Apr 21, 2012

### physical..O.o

Thanks for replying! =D but im not sure i know what u mean exactly, i realise that Pi=Pf but im not sure what to do with the angles you mentioned:

im guessing something to do with vector addition possibly taking in the angles? or with kinetic formulas?... could you explain a little more please?

6. Apr 21, 2012

### ehild

The momentum is a vector quantity. The overall momentum before collision is equal to that after collision. $\vec P_1 = \vec P_1' +\vec P_2'$
The addition holds for both the x and y components. The x component of the momentum P is magnitude times cos(angle), the y component is equal to the magnitude times sin(angle)

ehild

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7. Apr 21, 2012

### physical..O.o

i thought thats what i did...i used vecor addidtion like this (see attachments) but cant i just use pythag to get the initial momentum, then divide that by the 4WD's mass? because its a right angle..i tried your way with the x and y components but i just got the same answer as what i got initially, but im not sure if i did it right...

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8. Apr 21, 2012

### physical..O.o

i just had an idea, is this what you meant ehild...do you think that if i could find the extra amount of distance the 4WD would have travelled if it didnt crash into the car, then added that to the braking distance of 15m and then used the u= √(2mugs) rule that it might work? im just trying to work it out now...

9. Apr 21, 2012

### physical..O.o

:O i think this MIGHT be it!

If i add the vectors of 15m and 15.8 m i get :
unknown distance after crash if crash didnt happen= √(15^2 + 15.8^2)
= 21.79m
then if i add that to the initial braking distance:
= 21.79 + 15
= 36.79m
and use that in u= √(2mugs)
u= √(2*0.6*9.8*36.79)
u= 20.8ms-1
u= 74.9kmh-1
the speed is said to be in the 70's...so maybe this is correct?!? =D

10. Apr 22, 2012

### ehild

You are right, P1' and P2' really enclose a right triangle! I did not notice. So your solution is correct, and the way you found the original speed is ingenious!

If you encounter with similar problems to relate distance travelled and change of velocity, you can think of Work-Energy Theorem: Change of KE = work done. When it is friction doing work: 0.5 m( v22-v12)= -mgμs, and you can cancel m:

v12-v22=2gμs.

I have slightly different result for the original speed. It is because you rounded during the calculations. Keep more significant digits and round at the end.

ehild

11. Apr 22, 2012

### physical..O.o

haha thanks! XD
and il keep all that in mind