A problem about the laws of conservation: Bullet colliding with a wood block

[Platon7]

Homework Statement

A cannon is fired vertically into a wood block with a mass of 1.4 kg that is at rest directly above it. If the bullet has a mass of 21 g and a velocity of 210 m/s, how high will the block rise in the air after the bullet has penetrated it?

Relevant Equations

KE, PE

I have tried the following. Firstly, we can write the law of conservation of impulse
m1*v1 = m1*v'1 + m2*v'2, that is the first equation for the system. Then we can write the law of conservation of energy, like KE1 + KE2 = PE1 + KE'2, and then we can substitute formulas, but as far as I can see we will get 4 unknows, namely v'1, v'2, h and v''2 and that is the problem. I think about using some kinematics laws but don't know whether it's relevant.

 

[PeroK]

Why do you think mechanical energy is conserved in this case?

 

[Platon7]

I don't know, the system seems to be closed

 

[PeroK]

I don't know, the system seems to be closed

It is a closed system. But, that does not mean that mechnaical energy is conserved.

Consider two cars of equal mass and equal but opposite speed crashing into each other. The total momentum is zero before the collision (momentum being a vector) and zero afterwards. But, the total energy before the collision is not zero. Mechanical (kinetic) energy is lost in such a head-on collision.

 

[bob012345]

Is there any further description in the problem such as what exactly does it mean by penetration? Does the bullet stop imbedded in the block or go all the way through it according to the problem?

 

[Platon7]

Is there any further description in the problem such as what exactly does it mean by penetration? Does the bullet stop imbedded in the block or go all the way through it according to the problem?

As far as I understood, goes through and continues moving.

 

[Platon7]

It is a closed system. But, that does not mean that mechnaical energy is conserved.

Consider two cars of equal mass and equal but opposite speed crashing into each other. The total momentum is zero before the collision (momentum being a vector) and zero afterwards. But, the total energy before the collision is not zero. Mechanical (kinetic) energy is lost in such a head-on collision.

Yeah, I didn't think about that and it makes sense, because while penetrating almost for sure some energy was spent on heating the block, but then how am I suppoused to move further with solution?

 

[bob012345]

As far as I understood, goes through and continues moving.

Then could you put the whole problem statement here exactly as it was given to you. Leave out no details.

 

[Platon7]

Then could you put the whole problem statement here exactly as it was given to you. Leave out no details.

it is a full statement as I have written

 

[PeroK]

Yeah, I didn't think about that and it makes sense, because while penetrating almost for sure some energy was spent on heating the block, but then how am I suppoused to move further with solution?

Consider the collision between the bullet and the block. If the bullet is embedded in the block (which is what you are supposed to assume), then what does conservation of momentum tell you?

 

[Platon7]

Consider the collision between the bullet and the block. If the bullet is embedded in the block (which is what you are supposed to assume), then what does conservation of momentum tell you?

m1*v1 = m1*v'1 + m2*v'2, where m1 and m2 - are masses of bullet and block respectively, v1 - velocity of bullet before the collision, v'1 and v'2 - after

 

[PeroK]

m1*v1 = m1*v'1 + m2*v'2, where m1 and m2 - are masses of bullet and block respectively, v1 - velocity of bullet before the collision, v'1 and v'2 - after

It tells you more than that. Remember that $m_1$ is embedded in $m_2$. What does that tell you about $v'_1$ and $v'_2$?

 

[Platon7]

oh, yeah, v'1 = v'2 = u, then we can calculate u, but what's next step?

 

[PeroK]

oh, yeah, v'1 = v'2 = u, then we can calculate u, but what's next step?

Projectile motion?

 

[Platon7]

Projectile motion?

decelerated by g

 

[PeroK]

decelerated by g

What else!

 

[Platon7]

What else!

I don't know, rectilinear also, we know the velocity before penetrating

 

[PeroK]

I don't know, rectilinear also, we know the velocity before penetrating

There's only gravity acting on the block and bullet after the collision.

 

[Platon7]

oh, now I see, then we can find h using (v^2-v0^2)/(2*a), right?

 

[PeroK]

oh, now I see, then we can find h using (v^2-v0^2)/(2*a), right?

Yes, or your original idea to use conservation of energy for the projectile motion phase.

 

[Platon7]

I didn't really understand that, so you mean, m1*v1^2 = (m1+m2)*g*h. I'm not sure about the right part of equation

 

[PeroK]

I didn't really understand that, so you mean, m1*v1^2 = (m1+m2)*g*h. I'm not sure about the right part of equation

That equation ignores the collision. This problem has two parts:

a) a totally inelastic collision.

b) vertical projectile motion.

You have to tackle these two parts separately.

 

[Platon7]

do you mean (m1+m2)*g*h = (m1+m2)*u^2/2?

 

[PeroK]

do you mean (m1+m2)*g*h = (m1+m2)*u^2/2?

That looks right.

 

[Platon7]

yeah, because after dividing both parts for (m1+m2) we will get sth, that I have already mentioned above

 

[Platon7]

but when we assume that the bullet has penetrated and left the block, can we solve the problem?

 

[PeroK]

but when we assume that the bullet has penetrated and left the block, can we solve the problem?

You have to assume that the bullet is embedded in the block. If it passes through, you don't know the speed of the block.

 

[Platon7]

You have to assume that the bullet is embedded in the block. If it passes through, you don't know the speed of the block.

Okey, that you for help

 

[bob012345]

but when we assume that the bullet has penetrated and left the block, can we solve the problem?

That would be a different problem and more data would have been provided to solve it. I do think the problem statement by the author could have been clearer.