Momentum in a perfectly inelastic collision

[as2528]

Homework Statement

Two automobiles of equal mass approach an intersection. One vehicle is traveling with a velocity of 13.0 m/s toward the east and the other is traveling north with a speed of v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0° north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth?

Relevant Equations

arctan(fy/fx)=theta
m1viy+m2v2iy=(m1+m2)vfy

I calculated:arctan(fy/13.0)=55=>fy=18.566 m/s

Then I calculated, using the momentum equation:

m1viy+m2v2iy=(m1+m2)vfy=>

mv2i=2*m*18.566=>v2=37.132 m/s

I thought that because the cars were stuck together, the kinetic energy from the northbound car would be lost. So, the speed would have lowered after the accident. However, the answer is 18.566 m/s. Why do we only have to calculate the arctan? Would the speed not drop after the strike due to the mass being larger since the cars are stuck, and since momentum is conserved, the car would have been traveling at 37.132 m/s before the strike?

 

[TSny]

I calculated:arctan(fy/13.0)=55=>fy=18.566 m/s

The 55^o angle is the direction of the final velocity of the stuck-together cars. So,

$55^o = \arctan\left(\frac{v_{f_y}}{v_{f_x}}\right)$. But you used 13.0 m/s for ${v_{f_x}}$.

 

[as2528]

The 55^o angle is the direction of the final velocity of the stuck-together cars. So,

$55^o = \arctan\left(\frac{v_{f_y}}{v_{f_x}}\right)$. But you used 13.0 m/s for ${v_{f_x}}$.

Yes I did use the 13.0 m/s. Is that an error? The number calculated from that expression is the answer to the question, which was extremely surprising to me since I believed that the speed of the northbound car will have fallen at that point.

 

[TSny]

I did use the 13.0 m/s. Is that an error?

I think you're getting the correct answer without realizing why. Since momentum is conserved, the total momentum vector of the two cars before the collision equals the total momentum vector after the collision. So, if the final momentum of the system is at 55 ^o, the total initial momentum of the system is also at 55^o. So, the initial x and y components of momentum of the system statisfy $$\tan 55^o = \frac{P^{sys}_{i,y}}{P^{sys}_{i,x}}.$$ The x-component of the initial total momentum of the system is due to the first car. The y-component of the initial total momentum of the system is due to the second car. So, $$\tan 55^o = \frac{mv_{2i,y}}{mv_{1i,x}} = \frac{v_{2i,y}}{v_{1i,x}}.$$ So, when looked at this way, you can use the 13.0 m/s for the denominator $v_{1i,x}$. But note that the numerator is $v_{2i,y}$, not $v_{f,y}$. So, this equation will yield the initial velocity of the second car (which is what the question asks for). This is what you calculated even though you thought you were calculating the final y-component of the velocity of the system.

 

[as2528]

I think you're getting the correct answer without realizing why. Since momentum is conserved, the total momentum vector of the two cars before the collision equals the total momentum vector after the collision. So, if the final momentum of the system is at 55 ^o, the total initial momentum of the system is also at 55^o. So, the initial x and y components of momentum of the system statisfy $$\tan 55^o = \frac{P^{sys}_{i,y}}{P^{sys}_{i,x}}.$$ The x-component of the initial total momentum of the system is due to the first car. The y-component of the initial total momentum of the system is due to the second car. So, $$\tan 55^o = \frac{mv_{2i,y}}{mv_{1i,x}} = \frac{v_{2i,y}}{v_{1i,x}}.$$ So, when looked at this way, you can use the 13.0 m/s for the denominator $v_{1i,x}$. But note that the numerator is $v_{2i,y}$, not $v_{f,y}$. So, this equation will yield the initial velocity of the second car (which is what the question asks for). This is what you calculated even though you thought you were calculating the final y-component of the velocity of t

I think you're getting the correct answer without realizing why. Since momentum is conserved, the total momentum vector of the two cars before the collision equals the total momentum vector after the collision. So, if the final momentum of the system is at 55 ^o, the total initial momentum of the system is also at 55^o. So, the initial x and y components of momentum of the system statisfy $$\tan 55^o = \frac{P^{sys}_{i,y}}{P^{sys}_{i,x}}.$$ The x-component of the initial total momentum of the system is due to the first car. The y-component of the initial total momentum of the system is due to the second car. So, $$\tan 55^o = \frac{mv_{2i,y}}{mv_{1i,x}} = \frac{v_{2i,y}}{v_{1i,x}}.$$ So, when looked at this way, you can use the 13.0 m/s for the denominator $v_{1i,x}$. But note that the numerator is $v_{2i,y}$, not $v_{f,y}$. So, this equation will yield the initial velocity of the second car (which is what the question asks for). This is what you calculated even though you thought you were calculating the final y-component of the velocity of the system.

Got it, thanks! I didn't realize that I was calculating the initial velocity, not the final.

 

[TSny]

To make sure you're understanding the problem, I recommend that you also work the problem using your equations given below, which involve the final velocity components.

Relevant Equations:: arctan(vfy/vfx)=theta [edited to replace fy by vfy and fx by vfx]
m1viy+m2v2iy=(m1+m2)vfy

To use the first equation to obtain vfy, you will first need vfx. (It is not 13.0 m/s). Can you see a way to get vfx?

 

[as2528]

To make sure you're understanding the problem, I recommend that you also work the problem using your equations given below, which involve the final velocity components.To use the first equation to obtain vfy, you will first need vfx. (It is not 13.0 m/s). Can you see a way to get vfx?

Yes, I used m1vfx+m2v2fx = (m1+m2)vfx
So:
13*m+m*0=(m+m)*vfx
13m=2mvfx
vfx=13/2
vfx=6.5 m/s

Then
arctan(vfy/6.5)=55
vfy=tan(55)*6.5
vfy=9.3 m/s