# Momentum in a perfectly inelastic collision

• as2528
In summary: Thank you for the explanation and clearing that up!In summary, the conversation discusses the calculation of the final velocity of two cars stuck together after a collision. The 55 degree angle represents the direction of the final velocity, and using the conservation of momentum equation, it is determined that the initial velocity of the second car can be calculated using the 13.0 m/s value. This explains why the final velocity is 18.566 m/s and not lower as expected.
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?

as2528 said:
I calculated:arctan(fy/13.0)=55=>fy=18.566 m/s
The 55o 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}}##.

MatinSAR and as2528
TSny said:
The 55o 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.

as2528 said:
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 55o. 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.

MatinSAR, PeroK and as2528
TSny said:
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 55o. 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
TSny said:
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 55o. 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.

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.

as2528 said:
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 and MatinSAR
TSny said:
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

TSny

## What is a perfectly inelastic collision?

A perfectly inelastic collision is a type of collision in which the colliding objects stick together after impact and move as a single combined mass. In such collisions, the maximum amount of kinetic energy is lost, but momentum is conserved.

## How is momentum conserved in a perfectly inelastic collision?

In a perfectly inelastic collision, the total momentum of the system before the collision is equal to the total momentum of the combined mass after the collision. This can be expressed mathematically as: $$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$$, where $$m_1$$ and $$m_2$$ are the masses of the colliding objects, $$v_1$$ and $$v_2$$ are their velocities before the collision, and $$v_f$$ is the final velocity of the combined mass.

## Why is kinetic energy not conserved in a perfectly inelastic collision?

Kinetic energy is not conserved in a perfectly inelastic collision because some of the kinetic energy is converted into other forms of energy, such as heat, sound, and deformation energy. Only momentum is conserved in such collisions.

## How do you calculate the final velocity in a perfectly inelastic collision?

The final velocity of the combined mass after a perfectly inelastic collision can be calculated using the conservation of momentum. The formula is: $$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$, where $$m_1$$ and $$m_2$$ are the masses of the objects, and $$v_1$$ and $$v_2$$ are their velocities before the collision.

## Can a perfectly inelastic collision occur in all types of collisions?

No, a perfectly inelastic collision is a specific type of collision that occurs under certain conditions where the colliding objects stick together after impact. It is more common in macroscopic collisions involving objects that can deform and stick together, such as vehicles in a car crash. In microscopic or elastic collisions, objects do not stick together, and kinetic energy is conserved.

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