- #1

- 2

- 0

thanks,

smt

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter smt
- Start date

- #1

- 2

- 0

thanks,

smt

- #2

- 737

- 0

- #3

- 6

- 0

Now, work backwards from where the block is stopped by the spring. The force is a linear function of distance. Find out what this function is. Integrate the linear function (force) with distance to get the momentum of the block just after impact. To integrate with distance go as follows. Let F=force and p=momentum.

F=dp/dt=dp/dx*dx/dt=dp/dx*p/m=1/m*d(.5*p^2)/dx

p=sqrt(2*integral_over_distance(Fm)).

Then do a conservation of momentum balance. The initial momentum of the bullet minus the momentum of the block just after impact is the momentum of the bullet leaving the block.

If the momentum of the bullet leaving the block comes out negative, then the bullet remains lodged in the block and whoever asked the question made a mistake.

The mechanical energy difference can be calculated because you already have all masses and velocities.

- #4

- 86

- 0

Before the impact, the bullet's kinetic energy can be calculated from mass and speed.

At the instant of exit, the total energy is the exit kinetic energy + the energy stored in the spring (no other energy dissipation is mentioned in the problem).

The energy stored in the spring is the integral of force x shift (IIRC, force in a spring is proportional to shift, so the integral is pretty simple).

Therefore, calculate the exit kinetic energy as starting kinetic energy minus the energy stored in the spring, and then calculate the exit speed from the exit kinetic energy.

- #5

alphysicist

Homework Helper

- 2,238

- 2

You don't need to calculate the momentum.

Before the impact, the bullet's kinetic energy can be calculated from mass and speed.

At the instant of exit, the total energy is the exit kinetic energy + the energy stored in the spring (no other energy dissipation is mentioned in the problem).

I don't believe this is correct. In an elastic collision the kinetic energy is conserved, but in this collision mechanical energy is lost during the collision.

- #6

- 86

- 0

How about the following:

a) before the impact we easily know that the only momentum is

[\(\displaystyle ]

q_1 = m_1 v_1

[\(\displaystyle ]

and the only energy is

[\(\displaystyle ]

E_1 = 1/2 m_1 v_1^2

[\(\displaystyle ]

b) immediately after the impact, the bullet and the block start moving together with the same speed (can we assume this? there is no dimension given for the block) and momentum is conserved

[\(\displaystyle ]

(m_1 + m_2)v_t = m_1 v_1

[\(\displaystyle ]

but if this was true, don't we already know the total kinetic energy also?

c) from this moment on, momentum is not conserved anymore due to the spring which does an external force and changes the total momentum like

[\(\displaystyle ]

F=dp/dt=-kx

[\(\displaystyle ]

[\(\displaystyle ]

dp=-kxdt

[\(\displaystyle ]

[\(\displaystyle ]

\Delta p = -k \int xdt

[\(\displaystyle ]

but if we want to integrate in time, we need to know how long the bullet stays inside the block. We don't know that as a time interval, but we know that as a distance so we may need to switch the integral to dx?

d) assuming the previous points were correct, then we would calculate delta p which immediately gives the exit speed of the bullet. The speed of the block would also be the same (but immediately after it will start to change due to the spring now affecting only the block) so we would also immediately calculate the kinetic energies of both the bullet and the block, and the elastic energy at the exit time, and by subtracting all these from the initial E we would get the "lost" energy during the impact.

What troubles me is point b), since I really don't know if the assumption makes sense or not...\)\)\)\)\)\)\)\)\)\)\)\)

- #7

- 86

- 0

BTW, how do you use LaTeX math formatting in this board?

- #8

- 350

- 1

BTW, how do you use LaTeX math formatting in this board?

Use the beloved tex tags. Just click on anyone's tex and you'll see what it looks like. Here is [tex]F=ma[/tex] for example.

And in general the structure for invoking a different mode on these types of forums is [mode] hi I'm in a new mode now! [/mode] and it could be quotes, tex, images, urls...

- #9

alphysicist

Homework Helper

- 2,238

- 2

\(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyleYes, I thought that the exercise assumed no lost energy, and that by "internal energy" it actually meant only that stored into the block due to the spring, but maybe it really meant some energy lost during the impact (fracture or heating?).

How about the following:

a) before the impact we easily know that the only momentum is

[\(\displaystyle ]

q_1 = m_1 v_1

[\(\displaystyle ]

and the only energy is

[\(\displaystyle ]

E_1 = 1/2 m_1 v_1^2

[\(\displaystyle ]

b) immediately after the impact, the bullet and the block start moving together with the same speed (can we assume this? there is no dimension given for the block) and momentum is conserved

[\(\displaystyle ]

(m_1 + m_2)v_t = m_1 v_1

[\(\displaystyle ]

but if this was true, don't we already know the total kinetic energy also?

\)\)\)\)\)\)

In this collision the bullet and block never have the same speed. The bullet is slowing down, and the block is speeding up, but the bullet leaves the block long before they ever have a change to reach the same speed.

The collision is fast enough that at the end of the collision, the block has a certain speed, but we can consider the spring to have not had time to compress at all. (To be precise, the work done by the spring during the collision is negligible for the purpose of finding the speed of the block after the collision, so the spring's effect can be ignored during the collision time.)

\(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystylec) from this moment on, momentum is not conserved anymore due to the spring which does an external force and changes the total momentum like

[\(\displaystyle ]

F=dp/dt=-kx

[\(\displaystyle ]

[\(\displaystyle ]

dp=-kxdt

[\(\displaystyle ]

[\(\displaystyle ]

\Delta p = -k \int xdt

[\(\displaystyle ]

but if we want to integrate in time, we need to know how long the bullet stays inside the block. We don't know that as a time interval, but we know that as a distance so we may need to switch the integral to dx?\)\)\)\)\)\)

Here again we assume that the spring is not compressed at all while the bullet is inside. The spring compression is considered to take place completely after the collision is over.

It is true that an integral over the spring force has to be done; however, if instead of the work done by the spring, you use the potential energy of the spring, then that integral has already been done for you. (The integration for the work of the spring is how the potential energy formula was derived.)

So after the collision, the total quantity (kinetic energy + spring potential energy) is conserved for the block/spring system.\)\)\)\)\)\)\)\)\)\)\)\)

- #10

- 86

- 0

In this collision the bullet and block never have the same speed. The bullet is slowing down, and the block is speeding up, but the bullet leaves the block long before they ever have a change to reach the same speed.

The collision is fast enough that at the end of the collision, the block has a certain speed, but we can consider the spring to have not had time to compress at all. (To be precise, the work done by the spring during the collision is negligible for the purpose of finding the speed of the block after the collision, so the spring's effect can be ignored during the collision time.)

I believe we're reading the exercise in different ways... To me, the fact that there is no data about the size of the block made me assume that it's safe to consider the block like a point and its size negligible, but to you it leads to assume that also the time is negligible.

Here again we assume that the spring is not compressed at all while the bullet is inside. The spring compression is considered to take place completely after the collision is over.

"The block moves 5.00 cm to the right after impact." says the exercise. How do you read this?

I have read it that when the bullet leaves the block, the block has already moved 5cm, but it not necessarily still (it's not at the end of the spring compression yet).

Do you interpret it as instead the block starting to move when the bullet exits, so that the 5cm refers to the maximum spring compression some time after the exit?

- #11

- 86

- 0

Use the beloved tex tags. Just click on anyone's tex and you'll see what it looks like. Here is [tex]F=ma[/tex] for example.

And in general the structure for invoking a different mode on these types of forums is [mode] hi I'm in a new mode now! [/mode] and it could be quotes, tex, images, urls...

Thanks! I was using the Wikispaces formatting instead of the correct one for this forum!

Pity there is no edit button to go back and fix it...

- #12

alphysicist

Homework Helper

- 2,238

- 2

I believe we're reading the exercise in different ways... To me, the fact that there is no data about the size of the block made me assume that it's safe to consider the block like a point and its size negligible, but to you it leads to assume that also the time is negligible.

When it enters the block, the bullet is going at 400 m/s, and when it leaves it is going at 100 m/s. They don't give the size of the block, but just based on those speeds and a "reaonable" size for the block the collision is over fast.

But the real reason is that the spring has not yet compressed to any extent by the time the collision is over.

"The block moves 5.00 cm to the right after impact." says the exercise. How do you read this?

I have read it that when the bullet leaves the block, the block has already moved 5cm, but it not necessarily still (it's not at the end of the spring compression yet).

Do you interpret it as instead the block starting to move when the bullet exits, so that the 5cm refers to the maximum spring compression some time after the exit?

Yes, I definitely read 5cm as the maximum spring compression. The sentence states that after impact (the moment at which the bullet first touches the block) the block moves a distance of 5cm; since they don't specify anything else that means that is the total extent of motion after impact. If they had wanted the 5cm to be at the exit point they would have specified that.

(Also, if 5cm was the distance travelled during the collision and the collision time was not short, I think we would need to know a lot more. I haven't calculated it, but I believe momentum would not be conserved, since the impulse from the spring would be nonzero and large enough to affect the momentum of the block. The 5cm would not give us an energy directly, since energy is not conserved during the collision and we would not know the final stretch of the spring to get its final energy. I think the point of this problem is we make the normal "fast collision" assumption that the collision is over before there is any impulse from external forces, so that momentum is conserved.)

Last edited:

- #13

- 45

- 0

thanks,

smt

what is the answer for part a of the question is it 1.56m/s?

- #14

gneill

Mentor

- 20,922

- 2,866

what is the answer for part a of the question is it 1.56m/s?

Nope. The bullet will be moving much faster than that!

- #15

- 45

- 0

Nope. The bullet will be moving much faster than that!

really cause when i read this "the speed at which the bullet emerges from the block" meaning the speed of the bullet when it is in the block not when it first shot?

- #16

gneill

Mentor

- 20,922

- 2,866

really cause when i read this "the speed at which the bullet emerges from the block" meaning the speed of the bullet when it is in the block not when it first shot?

"emerges" = "exits from"

The bullet first encounters the block at 400 m/s, passes through the block, and finally emerges from the block at some lower velocity. Part (a) is asking for the speed of the bullet when it exits from the block (and it will carry on at that speed indefinitely).

Share: