A lead bullet gets embedded in a lead block

[hmparticle9]

Homework Statement

A lead bullet of mass $m$ and specific heat $c$ moving at $v_0$ gets embedded in a lead block of mass M. Assuming that the loss of kinetic energy goes into heating, what is the rise in temperature of the final object.

Relevant Equations

$\Delta Q = mc\Delta T$. $Q$ is heat, $m$ is mass and $T$ is temperature.

I have the following:

$$\Delta Q = (m+M) c \Delta T$$

and

$$K = \frac{1}{2}m v_0^2$$

I can not say that

$$\Delta Q = \frac{1}{2}m v_0^2$$

as the left is heat and the right is energy. Help required

 

[erobz]

Homework Statement: A lead bullet of mass $m$ and specific heat $c$ moving at $v_0$ gets embedded in a lead block of mass M. Assuming that the loss of kinetic energy goes into heating, what is the rise in temperature of the final object.
Relevant Equations: $\Delta Q = mc\Delta T$. $Q$ is heat, $m$ is mass and $T$ is temperature.

I have the following:

$$\Delta Q = (m+M) c \Delta T$$

and

$$K = \frac{1}{2}m v_0^2$$

I can not say that

$$\Delta Q = \frac{1}{2}m v_0^2$$

as the left is heat and the right is energy. Help required

Actually you can say that heat is change in kinetic energy here since nothing is assumed to have motion after the impact.

 

[renormalize]

Help required

All of your equations involve heat and kinetic energy. What about conservation of linear momentum?

 

[renormalize]

Actually you can say that heat is change in kinetic energy here since nothing is assumed to have motion after the impact.

Where does it explicitly state that the lead block is not moving after the collision?

 

[hmparticle9]

I seem to be getting the incorrect answer though. The answer in the back of the book is:
$$\Delta T = \frac{1}{2}\frac{mM}{(m+M)^2}\frac{v_0^2}{c}$$
What I get equating the change in kinetic energy to change in heat is:
$$\Delta T = \frac{1}{2}\frac{m}{(m+M)}\frac{v_0^2}{c}$$

what am i missing here?

 

[hmparticle9]

Nothing is said about the velocity after collision.

 

[hmparticle9]

If conservation of momentum has anything to say then the velocity of the system after collision is:

$$v = \frac{m }{m+M}v_0$$

 

[erobz]

Where does it explicitly state that the lead block is not moving after the collision

All of the kinetic energy is heat. If the lead block has no kinetic energy, it doesn’t have momentum either. Does it explicitly say it… no, but I think it implies it, perhaps unintentionally when it says “of the final object”

 

[renormalize]

I seem to be getting the incorrect answer though. The answer in the back of the book is:
$$\Delta T = \frac{1}{2}\frac{mM}{(m+M)^2}\frac{v_0^2}{c}$$

The book answer is correct if you assume that the velocity of the lead block is zero before the bullet collides with it. Just apply conservation of momentum!

 

[hmparticle9]

First, just so I know, what was I doing wrong? and second I don't see how the answer comes from conservation of momentum.

 

[renormalize]

All of the kinetic energy is heat.

You've made an invalid assumption. The problem states "Assuming that the loss of kinetic energy goes into heating...". Nowhere does it say that all the kinetic energy goes into heat.

 

[erobz]

You've made an invalid assumption. The problem states "Assuming that the loss of kinetic energy goes into heating...". Nowhere does it say that all the kinetic energy goes into heat.

Ok, fair point. I think I was under -idealizing.

 

[renormalize]

First, just so I know, what was I doing wrong? and second I don't see how the answer comes from conservation of momentum.

Assuming the initial velocity of the lead block is zero, we have:
Conservation of energy:$$\frac{1}{2}m v_0^2 = \frac{1}{2}(m+M) v^2+\Delta Q$$where $v$ is the velocity of block+bullet after collision.
Conservation of momentum:$$m v_0=(m+M)v$$That's 2 equations with 2 unknowns $v,\Delta Q$. Solve for both unknowns and use $\Delta Q$ to find $\Delta T$.

 

[hmparticle9]

Thank you :)

 

[erobz]

So now that the doubt is clear; some idealizations of the “final state” ( none of which have been explicitly stated “ into heating” - what?) are 1) “immediately” after collision 2) when the system comes to rest, and 3) the heat death of the universe. All different answers?

 

[haruspex]

So now that the doubt is clear; some idealizations of the “final state” ( none of which have been explicitly stated “ into heating” - what?) are 1) “immediately” after collision 2) when the system comes to rest, and 3) the heat death of the universe. All different answers?

I agree, the problem is quite ambiguous.
Immediately after impact, the whole object doesn’t have "a temperature". The bullet and immediate surroundings will be a lot hotter than the block in general.
How will the conglomerate be moving by the time temperatures have more-or-less evened out? Depends on the setting. Maybe the system is in space, or falling from a great height, or the block is swinging from a cable… Or maybe the block was resting on the ground and has long since come to rest because of friction which has turned the remaining KE into.. oh, heat.

 

[Steve4Physics]

Getting only slightly off topic, here is a challenge for anyone so inclined...

Is the following a complete/sufficient problem statement? I believe so. Or is there anything (non-trivial) not addressed?
_________

A lead block of mass $M$ is at rest on a frictionless horizontal surface. A lead bullet of mass $m$ and speed $v_0$ strikes and becomes embedded in the block.

The bullet’s path is a horizontal straight line towards the block’s centre of mass, so the block+bullet translates without rotation.

The block and bullet were at the same initial temperature. Take the specific heat capacity of lead as $c$ and assume there is negligible heat-loss from the block+bullet to the surroundings.

What is the final equilibrium temperature-rise of the block+bullet?

Edit to add the word 'rise'.

 

[erobz]

Getting only slightly off topic, here is a challenge for anyone so inclined...

Is the following a complete/sufficient problem statement? I believe so. Or is there anything (non-trivial) not addressed?
_________

A lead block of mass $M$ is at rest on a frictionless horizontal surface. A lead bullet of mass $m$ and speed $v_0$ strikes and becomes embedded in the block.

The bullet’s path is a horizontal straight line towards the block’s centre of mass, so the block+bullet translates without rotation.

The block and bullet were at the same initial temperature. Take the specific heat capacity of lead as $c$ and assume there is negligible heat-loss from the block+bullet to the surroundings.

What is the final equilibrium temperature-rise of the block+bullet?

Edit to add the word 'rise'.

I feel it is complete(almost).

Edit: Maybe drag needs ignored too, but much closer to complete than the OP. It’s funny how wordy it will become to be clearly stated ( it isn’t trivial wording change ). I had to read it a few times, keeping a mental checklist…then still drag popped in to my head as warming the system with the bullets initial kinetic energy.