What is the final velocity of the car in an elastic collision?

[Hypnos_16]

Homework Statement

A 1056-kg van, stopped at a traffic light, is hit directly in the rear by a 705-kg car traveling with a velocity of +2.45 m/s. Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of the car?

Van
m = 1056kg
v = 0
p = 0

Car
m = 705
v = 2.45
p = 1727.3

Homework Equations

i know that in this kind of collision problem the momentums would add
mv + mv = mv
the mass and velocities of the two vehicles would add because they would both wind up going the same way, the trouble I'm having is isolating just the speed of the car.

The Attempt at a Solution

mv + mv = mv
(1056)(0) + (705)(2.45) = (1761)v
0 + 1727.3 = 1761v
1727.3 = 1761v
v = 0.981m/s

 

[Doc Al]

and the collision is elastic.

If the collision is elastic, the vehicles do not stick together after the collision. (Unless this is a misprint, which I suspect it is.)

 

[Hypnos_16]

okay, suppose it is a misprint, and that the two cars do stick to one another, how would i go about isolating juts the speed of the car?

 

[Doc Al]

okay, suppose it is a misprint, and that the two cars do stick to one another, how would i go about isolating juts the speed of the car?

If they stick together, then they have the same speed. Just like you calculated.

 

[Hypnos_16]

then it must not be a misprint, because the answer i got there isn't the correct answer, i guess they don't stick together.

 

[Doc Al]

then it must not be a misprint, because the answer i got there isn't the correct answer, i guess they don't stick together.

In that case you need to redo your conservation of momentum equation. But that's not enough--what else is conserved in an elastic collision?

 

[Hypnos_16]

Momentum and Energy are Conserved. so if momentum in conserved would pf = pi ?

 

[Doc Al]

Momentum and Energy are Conserved.

Good.

so if momentum in conserved would pf = pi ?

Yes.

 

[Hypnos_16]

so then, since his mass didn't increase, his final velocity should be the same has his initial velocity? that doesn't seem to make sense

 

[Doc Al]

so then, since his mass didn't increase, his final velocity should be the same has his initial velocity? that doesn't seem to make sense

The total momentum is conserved, not the momentum of each car separately.

 

[Hypnos_16]

But in this case wouldn't total momentum be juts the value of the car's momentum since it's the only thing moving?

∑p = p1 + p2
∑p = (705)(2.45) + 0
∑p = 1727.3

then wouldn't v then just be 2.45m/s again?

 

[Doc Al]

But in this case wouldn't total momentum be juts the value of the car's momentum since it's the only thing moving?

Yes.

∑p = p1 + p2
∑p = (705)(2.45) + 0
∑p = 1727.3

OK.

then wouldn't v then just be 2.45m/s again?

No. After the collision you'll have two speeds to worry about.

 

[Hypnos_16]

So if there are now two speeds to work, would both speeds be the same?
∑p = p1 + p2
∑p = (705)(2.45) + 0
∑p = 1727.3

just divide the momentum in two, half going to one vehicle and the other half going to the other vehicle

so
Car = p = mv
863.7 = (705)v
v = (1.23m/s)?

 

[Doc Al]

So if there are now two speeds to work, would both speeds be the same?

No.

just divide the momentum in two, half going to one vehicle and the other half going to the other vehicle

No.

You must apply conservation of momentum and energy to solve for the speeds.

 

[Hypnos_16]

But how can i solve for two variables if i know nothing about their speeds after the collision?

 

[Doc Al]

But how can i solve for two variables if i know nothing about their speeds after the collision?

The speeds after the collision are what you are trying to find. You'll have two equations and two unknowns. Just right.

 

[Hypnos_16]

Okay i think i screwed something up,
i have the ∑p = 1727.3
so i use that as Pf = Pi
Pi = 1727.3
so that what?
1727.3 = mv?
and just input the masses one at a time back into m?

 

[Doc Al]

Okay i think i screwed something up,
i have the ∑p = 1727.3
so i use that as Pf = Pi
Pi = 1727.3
so that what?

That's the total momentum.

1727.3 = mv?
and just input the masses one at a time back into m?

No. Use the full version of the conservation of momentum equation:

$$m_1\vec{v}_1= m_1\vec{v}_1{ '} + m_2\vec{v}_2{ '}$$

You'll also need an equation for conservation of energy.