Connected system of a bullet and a block

[vxr]

Homework Statement

A bullet of mass $m = 0.1 kg$ embeds itself in a block of mass $M = 10kg$, which is attached to a spring of elastic force constant $k = 10 \frac{N}{m}$. If the initial velocity of bullet is $v_{0} = 1 \frac{m}{s}$, determine the maximum compression of a spring $x$ and the time $t$ for which the bullet-block system comes to rest

Relevant Equations

I believe these can be useful: $F = -kx, \quad U = \frac{kx^2}{2}, \quad E_{k_{0}} = \frac{(m+M)v_{0}^2}{2}$

This is task from my textbook and it does not provide us with an answer. So I cannot verify if I did mistake. Can someone double check, please? My solution:

$E_{k_{0}} = \frac{(m+M)v_{0}^2}{2} \quad \land \quad U = \frac{kx^2}{2}$

$E_{k_{0}} = U$

$\Longrightarrow (m+M)v_{0}^2 = kx^2$

$x = \sqrt{\frac{(m+M)v_{0}^2}{k}} = \sqrt{\frac{m + M}{k}} |v_{0}|$ (the result is positive, should it be positive?)

--

and now the $t$:

$F = -kx$

$ma = -kx$

$a = -\frac{kx}{m}$

$\frac{\Delta v}{t} = - \frac{kx}{m}$

$\frac{-v_{0}}{t} = - \frac{kx}{m}$

$t = \frac{mv_{0}}{kx}$ (plug in calculated x from above equation here)

Not sure if this result is good and if in the time equation the $m$ should be $m$, $M$ or $(m + M)$.

Help is appreciated, thank you.

 

[BvU]

Hi,

$E_{k_{0}} = U$

What makes you think this applies ?
Also, initially the block is at rest, so why give it kinetic energy corresponding to $v_0$ ?

 

[vxr]

No idea. I have just assumed that it indeed does apply here. Can I get any advice how should I find the $x$?

 

[BvU]

embeds itself in a block

is a characteristic of a fully inelastic collision
Kinetic energy is not conserved. What is ?

 

[vxr]

Momentum of the system?

 

[jbriggs444]

Momentum of the system?

Yes. The typical assumption is that a collision is brief enough that any other external forces during the collision produce negligible changes in momentum.

Given that momentum is conserved for the collision event, what can you calculate next?

 

[vxr]

Well, if momentum is conserved, then I assume:

Bullet's momentum:

$p_{1} = mv_{0}$

Resting block's momentum:

$p_{2} = 0$

Connected system's initial momentum:

$p = p_{1} + p_{2}$

$(m+M)v = mv_{0}$

$v = \frac{mv_{0}}{m+M}$

But what I need to find is $x$ and $t$.

So if I know the $v$, I can use the equation

$E_{k} = U$ now and find $x$? Or this equation still does not apply? If not, then how can I find $x$?

Also what about my calculation of $t$, is it correct (assuming that I plug in proper $x$ value in there)?

Thank you!

 

[jbriggs444]

Well, if momentum is conserved, then I assume:

Bullet's momentum:

$p_{1} = mv_{0}$

Resting block's momentum:

$p_{2} = 0$

Connected system's initial momentum:

$p = p_{1} + p_{2}$

$(m+M)v = mv_{0}$

$v = \frac{mv_{0}}{m+M}$

Very nice.

So if I know the $v$, I can use the equation

$E_{k} = U$ now and find $x$?

Bingo. With the collision over, there are no further energy losses to worry about. Conservation of energy can get you the displacement.

Also what about my calculation of $t$, is it correct (assuming that I plug in proper $x$ value in there)?

It looks like your calculation of t assumes constant acceleration. I think that you need a different approach. Have you studied simple harmonic motion?

It could be tempting to guess that average acceleration is given by half the maximum acceleration. But that approach would be flawed. You would be effectively using a distance-weighted average. You need a time-weighted average if you are planning to divide velocity by average acceleration to yield time.