Collision of Car & Truck: Calculating Velocity & Direction

[BilloRani2012]

A car weighing m kg, and a truck, weighing 4m kg, collide at an angle of alpha. Initially they are traveling at the same velocity. After the collision they move off together at an angle of theta.

a) show that tan(theta) = (4 sin alpha) / (4 cos alpha + 1)

b) Show that if their initial velocity is u the velcoity that they move off at is
v = u/5 squareroot of (17+8 cos alpha)

c) Find the velocity and direction they move off at if alpha = pi/2. What value of alpha gives the minimum velocity?

 

[tiny-tim]

Welcome to PF!

Hi BilloRani2012! Welcome to PF!

(have an alpha: α and a theta: θ and a pi: π and a square-root: √ )

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[BilloRani2012]

i've proved part a)

i just don't know what to do for part b) an c)

 

[tiny-tim]

Hi BilloRani2012 !

Show us your full calculations so far, and then we'll see what went wrong, and we'll know how to help!

 

[anny_cool]

wieght is a...scalar or vector quantity
area...
volume...
length
pressure

 

[tiny-tim]

welcome to pf!

hi anny_cool! welcome to pf!

if that's a question, please start a new thread for it (and show what work you've done so far)

 

[BilloRani2012]

This is what I've done so far:

a) tan theta = 4m*sin alpha / m + 4m cos alpha
therefore tan theta = 4 sin alpha / 4 cos alpha + 1

b)m = car and 4m = truck
so, 4m*u + m*u = 5mu

tan theta = o/a = 4 sin alpha / 4 cos alpha + 1

then I'm stuck from there...

 

[tiny-tim]

Hi BilloRani2012!

(just got up :zzz: …)

a) tan theta = 4m*sin alpha / m + 4m cos alpha

(what happened to that α and θ i gave you? )

how did you get that?

(if you got it the way you should have, then (b) should be easy)

 

[BilloRani2012]

i still don't know how to do b)

 

[tiny-tim]

Please show how you got a) …

a) show that tan(theta) = (4 sin alpha) / (4 cos alpha + 1)

 

[BilloRani2012]

okay..

By the law of conservation:

on x axis
--> m + 4m*cos alpha = (m+4m)*cos theta --------- Equation 1

On y axis
--> 0 + 4m*sin alpa = (m+4m)*sin theta ------------ Equation 2

So, divide equation 2 by equation 1:

tan theta = (4m*sin alpa) / (m + 4m cos alpha)

Therefore tan theta = 4 sin alpha / 4 cos alpha + 1

 

[tiny-tim]

By the law of conservation:

on x axis
--> m + 4m*cos alpha = (m+4m)*cos theta --------- Equation 1

On y axis
--> 0 + 4m*sin alpa = (m+4m)*sin theta ------------ Equation 2

ah, now I see the difficulty …

conservation of what?

momentum?

then you'll need to put velocities into your equations …

once you've done that, you should be able to see how to get b) ​

 

[BilloRani2012]

yes conservation of momemtum.

Um i don't really get it. So do i have to replace m with v??

 

[tiny-tim]

momentum before is mu, and after is mv

(and yeeees … a momentum equation must have momentum in it! )

 

[BilloRani2012]

okay...

so would the equation be:

mv=mu
tan theta = 4mu/4mu + 1

so just replace sin alpha and cos alpha with mu??

 

[tiny-tim]

No.

The RHS of your Equation 1 …

--> m + 4m*cos alpha = (m+4m)*cos theta --------- Equation 1

… should be (m+4m)*v*cos theta …

so what should the LHS side be?

(and what should Equation 2 be?)​

 

[BilloRani2012]

okay,

so my Equation 2 was:
0 + 4m*sin alpa = (m+4m)*sin theta ------------ Equation 2

so should it be:
0 + 4m*u = (m+4m)*u

is that right??

 

[BilloRani2012]

sorry i made a mistake. Would the LHS of EUATION 1 be this:

m + 4m*v*cos alpha = RHS