Question involving conservation law and springs

[mizzy]

Homework Statement

A freight car of mass 1800kg is timed at 4.2m/s just after it runs into a spring-loaded bumper at the end of the track. The bumper consists of an 800kg mass that the car runs into, and a pair of large springs. The car travels 2.4m before coming to rest.

a)using an appropriate conservation law, find the speed of the freight car before it struck the bumper.

Homework Equations

i'm not sure, conservation of momentum? or kinetic energy?

The Attempt at a Solution

Before it struck the bumper, it's moving at a certain speed(unknown) at a given mass. hmmm...looking at what's given to find the initial speed we have to use conservation of momentum: m1v1 +m2v2 = 1800v1 + 800(4.2)

Does that look right? Can someone guide me please?

 

[kuruman]

It is indeed conservation of linear momentum, but you are not applying it correctly.

You need to say P_before = P_after.

Can you find expressions for P_before and P_after?

 

[mizzy]

ok. so p before = p after

m1v1 + m2v2 = m1v1 + m2v2

It's an elastic collision, right?

i calculated it and i got speed before = 10.9m/s

There's a second part to this asking for the time taken for the car to be brought to rest. do i have to consider the springs?? or can i just use kinematics equations

 

[kuruman]

ok. so p before = p after

m1v1 + m2v2 = m1v1 + m2v2

This equation is an identity. Since the terms on each side are the same it basically says 0 = 0. Can you show me how you got 10.9 m/s?

We will worry about the second part after it is certain that you got the first part correctly.

 

[mizzy]

This equation is an identity. Since the terms on each side are the same it basically says 0 = 0. Can you show me how you got 10.9 m/s?

We will worry about the second part after it is certain that you got the first part correctly.

pbefore = pafter

m1v1 +m2v2 = m1v1 + m2v2
1800v1 = 1800 *4.2 +800v2
1800v1 = 7560 +800v2

using conservation of energy:
v1i - v2i = -v1f + v2f
v1i - 0 = -4.2 + v2f

solve for v2f. v2f = v1i + 4.2

1800v1 = 7560 + 800(v1i + 4.2)
1800v1 = 7560 + 800v1 + 3360
1000v1 = 10920
v1 = 10.9m/s

 

[kuruman]

pbefore = pafter

m1v1 +m2v2 = m1v1 + m2v2
1800v1 = 1800 *4.2 +800v2

Picture the situation right after the collision: The car pushes on the bumper which pushes on the springs that get compressed until the car stops. If the car is moving initially at 4.2 m/s just after the collision, what do you think is the speed of the bumper?

using conservation of energy:
v1i - v2i = -v1f + v2f

energy is not conserved because the masses stick together and stay stuck. Besides, this is not an expression of energy conservation. What is the equation for kinetic energy? Check your textbook.

 

[mizzy]

the speed of the bumper would b 4.2m/s?

This is an inelastic collision then?

m1v1 + m2v2 = (m1 + m2)v2
1800v1 = (1800 + 800) 4.2
v1 = 2600/1800
v1 = 1.44m/s

 

[kuruman]

the speed of the bumper would b 4.2m/s?

This is an inelastic collision then?

Yes.

m1v1 + m2v2 = (m1 + m2)v2

Correct, but to avoid confusion, use one symbol to denote one and only one variable. Here v2 stands for the velocity of the bumper before the collision (which is zero) and for the common velocity of the masses after the collision (which is 4.2 m/s). This habit might get you in trouble in the future.

1800v1 = (1800 + 800) 4.2

Correct.

v1 = 2600/1800
v1 = 1.44m/s

What happened to the "4.2" in the previous equation?

 

[mizzy]

Yes.

Correct, but to avoid confusion, use one symbol to denote one and only one variable. Here v2 stands for the velocity of the bumper before the collision (which is zero) and for the common velocity of the masses after the collision (which is 4.2 m/s). This habit might get you in trouble in the future.

Correct.

What happened to the "4.2" in the previous equation?

ooops, i forgot to multiply 4.2. The answer i get is 6.01m/s.

 

[kuruman]

Good. For the second part, you cannot use the kinematic equations because they are good for constant acceleration. Masses at the ends of springs do not experience constant acceleration. The quickest way to get the second part is energy conservation.

 

[mizzy]

So using PEg + PEs + KE = PEg + PEs + KE

but we need to find time.

How do we do that?

 

[kuruman]

So using PEg + PEs + KE = PEg + PEs + KE

but we need to find time.

How do we do that?

What fraction of the period T is the time that you are looking for?

 

[mizzy]

What fraction of the period T is the time that you are looking for?

I don't quite understand what you mean about what fraction? All i know is that the car travels 2.4m before coming to rest. Do i have to find the final velocity and find time from that?

 

[kuruman]

When it comes to rest, does it remain at rest or is it instantaneously at rest and then goes back where it came from?

 

[mizzy]

it goes back and forth since the bumper is attached to the spring

 

[kuruman]

OK, in that case it has a period T which is the time required for a complete round trip. Here the masses go from the point of collision to maximum compression of the springs, which is not a round trip, just a piece (fraction) of it. What piece (fraction) of the period T then is the corresponding time it takes the masses to go from the point of collision to the maximum compression of the springs?

 

[mizzy]

we want half of the period...is that right?

 

[kuruman]

A period is the time for a round trip. Therefore

From maximum stretching of the springs to maximum compression it is ______ of a period. So ...

 

[mizzy]

isnt' it half? because one period would be from stretch to compression to stretch...

sorry if I'm not catching on right away =(

 

[kuruman]

Half a period is from maximum stretch to maximum compression. When the car hits the block, the springs are relaxed, not at maximum stretch. So ...

 

[mizzy]

quarter?? ok...i'm guessing and that's not good

 

[kuruman]

No reason to guess. Draw yourself a sine wave representing the motion of the masses over a complete period starting at zero displacement. You have one "mountain" and one "valley". Maximum compression is from the starting zero to the top of the mountain. What fraction of the total is that?

 

[mizzy]

1/4th

 

[kuruman]

Correct. Can you find the frequency of oscillations ω? If so how is it related to the period T?

 

[mizzy]

omega = 2pi*f

therefore, f= omega/2pi

 

[kuruman]

And how is f related to T? Also, can you find an expression for f in terms of the masses and the spring constant?

 

[mizzy]

The inverse of the period is the frequency.

the expression for f with terms of masses and spring constant is

f = 1/2pi square root k/m

 

[kuruman]

Good. Now can you put it together? Find what T is in terms of k, m and pi. The time that you are looking for is 1/4 of that.

 

[mizzy]

k. T = 2pi square root m/k (k is equal to mg/x, where x is the distance of 2.4m?)

the question asks for the time is took the car to be brought to rest. So i know we came up with that the time is a quarter of the period and that is at max compression. so how is that when the car is at rest?

 

[kuruman]

k. T = 2pi square root m/k (k is equal to mg/x, where x is the distance of 2.4m?)

No. k = mg/x applies to vertical springs that stretch by amount x when weight mg is hung on them. The spring constant k should be given by the problem. If it is not, then you cannot find the period. Can you post the statement of the problem exactly as it is given?

the question asks for the time is took the car to be brought to rest. So i know we came up with that the time is a quarter of the period and that is at max compression. so how is that when the car is at rest?

What does maximum compression mean? It means that the mass has traveled as far in as possible. This means that it stops (instantaneously) because if it didn't, it would still move and you would not maximum compression.