Impulse/momentum Car collision

[Jason03]

Heres the problem I am working on

http://img91.imageshack.us/img91/1161/cartz0.jpg [Broken]

Im trying to figure the setup to find which car was going faster and how fast...

Would I set up the equation before the collision and after...the collision is perfectly plastic so

$$m_{a}v_{a} + m_{b}v_{b} = (m_{a} + m_{b})v\prime$$

also since its perfectly plastic

$$v\prime_{b} = v\prime_{a} = v\prime$$

 

[rock.freak667]

$$m_{a}v_{a} + m_{b}v_{b} = (m_{a} + m_{b})v\prime$$

This would be true if A and B were both traveling in the same direction.

One travels horizontally (horizontal momentum) and the other travels vertically (vertical momentum). The components of the final momentum should be equal to those two.

 

[Jason03]

ok, I understand breaking up the momentum into components, but what about finding the Velocity?

Since I am trying to find which car was going faster and how fast, I only know that one of the cars was going 35 mi/h...

 

[rock.freak667]

Post the equations you got.

If you could find a way to compare the two initial velocities you should be able to deduce which car was traveling faster.

 

[Jason03]

im having trouble setting up the components of momentum,...I understand that one is vertical and one is horizontal...So do I have 2 different sets of x and y equations?,,...or can I include it all in one equation?

Since car B is horizontal would I just do V_b cos 0...and for car A v_a sin 90?

im not sure...

 

[rock.freak667]

Conservation of linear momentum: Initial momentum=Final momentum

Horizontally:

$m_B v_B=(m_A+m_B)Vcos40$

Vertically:

$m_Av_A=(m_A+m_B)Vsin40$

 

[Jason03]

Ok that helps...I tried plugging in the 35mi/hr(51 ft/s) for the initial velocities in each set of equations and solving for the final V...but I am not sure if that is an accuarte way to find who was going faster and how fast...

I know the answer is b and that b was going 52.9 mi/h...but still trying to get there...

 

[rock.freak667]

hmm...I thought it was car A that was going faster. But anyhow, seeing as how I have the first part wrong, I can't really help you with that.

But for the 2nd part, you would use the equations as you have correctly done.

 

[Jason03]

yea i spoke to some other people from my class and they haven't been able to get the answer either...class is Tuesday night...so I will keep trying...and thanks for your help

 

[rohanprabhu]

Clearly, there are three unknowns [velocity of car A, velocity of car B and the final velocity]. And we have two equations only [a vector component for each of the directions]. Hence it is impossible to solve for these values. What is possible, however is to solve for their ratios.

Let the velocity and mass of car A be $v_a$ and $m_a$ and the velocity and mass of car B be $v_b$ and $m_b$.

Initial momenta of the cars are given as:

$$\overrightarrow{p}_a = m_a v_a \hat{j}$$

and

$$\overrightarrow{p}_b = m_b v_b \hat{i}$$

Hence the initial momentum of the system [momentum just before the collision] is:

$$\overrightarrow{p} = \overrightarrow{p}_a + \overrightarrow{p}_b = m_b v_b \hat{i} + m_a v_a \hat{j}$$

Let the magnitude of the velocity just after the collision of the system be: $v_f$. Then, the final momentum just after collision is given as:

$$\overrightarrow{p}_f = (m_a + m_b)v_f(\cos(\theta) \hat{i} + \sin(\theta) \hat{j})$$

On equating the two momenta, we get:

$$m_b v_b \hat{i} + m_a v_a \hat{j} = (m_a + m_b)v_f(\cos(\theta) \hat{i} + \sin(\theta) \hat{j})$$

On equating the $\hat{i}$ and the $\hat{j}$ components, we get:

$$m_b v_b = (m_a + m_b)v_f \cos(\theta)$$

$$\frac{v_b}{v_f} = \frac{(m_a + m_b)}{m_b}\cos(\theta)$$ (equation 1)

similarly, for the $\hat{j}$ component:

$$\frac{v_a}{v_f} = \frac{(m_a + m_b)}{m_a}\sin(\theta)$$ (equation 2)

Dividing equation 1 by equation 2, we get:

$$\frac{v_a}{v_b} = \frac{m_b}{m_a} \tan(\theta)$$

Substitute the values and you can come to know by the $v_a / v_b$ ratio, which velocity is higher. Then substitute, the value of 35 mph for the lower one and u shall get the higher velocity too.

 

[Rayquesto]

well based on the problem, I believe that if each car were to go the same speed, then the angle they would form would be 45 degrees, but due to the fact that its 40 degrees, then that means that the car going east is going about cos5 faster than the car going north, car A going north would have a change in theta of -5, Car b would have a chnage in theta of 10, where a perfectly inelastic collision occured, and that one car was 2000 lb and the other 3600 lb, then the equation should look something like this: cos40vf(5600lb)=3600lbvicos(10) plus 2000lbvisin(-5) i'M NO expert since I am only in physics A, but this is what I would do since if we take wo cars going in a 90 degree direction forming a resultant vector going 40 degrees. this will simplify to 3545.3via-174.3vib=4290vf which. So, I think that Car B was going faster. finding either of initial velocity or final velocity would both be extremely tough and i don't really have any ideas, but I suppose you can plus in 35mph to make either A or B valid.

 

[karalyn84]

No physics pro here, was wondering if anyone can tell me how many mph its possible for a car to go in 28 feet. jeep grand cherokee. ballpark. Thank you so very much i greatly appreciate it! ~KK

 

[Rayquesto]

i think that will depend upon the mass of the car, and the power of the engine. I am pretty sure that once you know the power, you can figure out the force the engine applies. once you know the force just divide it by the mass of the car. once you know the acceleration, you can use some equations to find mps, not mph yet.

28ft=8.53439999meters. so 8.53meters=v0t(this is zero if the car starts at rest even with the engine on) plus 1/2at^2 *you don't know time

or use

Vf^2=Vi^2(probably zero) plus 2(acceleation)(distance) since you don't know how much tim it will go but you do know the acceleration from the engine power info.

once you find vf, you can divide that number by 3,600 to get ft/h

 

[Rayquesto]

sorry to get m/hr

 

[Rayquesto]

meters/hour