The velocity of a movable block that is penetrated by an arrow

[PrincessPeach]

Homework Statement

an arrow of mass 50g is fired horizontally from short range into a target consisting of a slap of expanded polystyrene glued between two blocks of wood. The mass of the target is 4 kg. The arrow remains embedded in the polystyrene, and the whole assembly slides a distance of 86mm along the steel table on which the target is resting. The coefficient of kinetic friction between the wood and the table is 0.3. Afterwards, it is determined that the arrow penetrated the polystyrene by 56mm.

Determine the velocity of the target immediately after impact. (Answer should be 1.03m/s)
Determine The velocity of the arrow immediately before striking the target. (Answer should be 83.43 m/s)

Relevant Equations

Momentum, Friction, and Energy

This problem is in a chapter on momentum in the book basic engineering mechanics explained.

Help me Mario

 

[PeroK]

According to the homework guidelines you have to post your best attempt at solving this problem yourself.

 

[PrincessPeach]

I apologize.

I tried to solve by using the conservation of energy where I had the kinetic energy of the arrow before impact equal the work required to penetrate the arrow (F . x1) & move the block (F . x2).

Then I tried to arrange the expression and solve the velocity of the arrow although I did not get the correct answer.

 

[PeroK]

I tried to solve by using the conservation of energy

Why do you think energy is conserved?

 

[PrincessPeach]

Why do you think energy is conserved?

I assumed that the kinetic energy from the arrow would be used to penetrate the block and move the block.

I saw a similar method performed in a similar question in the link below (Hint 3 - Eqn (3)).

https://physicstasks.eu/1994/bullet

 

[PeroK]

I assumed that the kinetic energy from the arrow would be used to penetrate the block and move the block.

I saw a similar method performed in a similar question in the link below (Hint 3 - Eqn (3)).

https://physicstasks.eu/1994/bullet

Perhaps you should give up on that site. This is what's called a totally inelastic collision, where kinetic energy is not conserved. Momentum, on the other hand, is conserved.

 

[PrincessPeach]

Okay, thank you for the insight.

I'm not sure as to how to use momentum to solve the problem when the velocity of the arrow or the target is not given in the question tbh.

 

[PeroK]

Okay, thank you for the insight.

I'm not sure as to how to use momentum to solve the problem when the velocity of the arrow or the target is not given in the question tbh.

You have to work out the velocity of the block after the collision from the other information.

 

[PrincessPeach]

You have to work out the velocity of the block after the collision from the other information.

I've tried to use Newton's second law to determine the deceleration generated by the resisting friction force.

Then using kinematic equations to solve for the initial velocity - 0.711m/s2.

Would you have another in mind?

 

[bob012345]

Has anyone been able to match the book answer of 1.03 m/s? I used the Work-Energy Theorem and also got 0.711 m/s.

 

[PeroK]

That looks like a good start. Now, can you use conservation of momentum to get the velocity of the arrow before the collision?

See below. Your answer is correct and the book is wrong.

 

[PeroK]

PS The book answer looks wrong to me. Note first, that the mass of the system is irrelevant:
$$F_f = \mu Mg \ \Rightarrow \ a = -\frac {F_f}{M} = -\mu g = -0.3 \times 9.8 m/s^2 = -2.94 m/s^2$$Now, suppose the book answer is correct and the initial velocity of the block is about $1m/s$. The motion will last about $0.34s$, at an average speed of $0.5m/s$, which gives a distance of about $0.17m$.

My conclusion is that whoever did the book solution did the following erroneous calculation:
$$v = 2\sqrt{as}$$You might need a new book as well!

 

[PrincessPeach]

That looks like a good start. Now, can you use conservation of momentum to get the velocity of the arrow before the collision?

See below. Your answer is correct and the book is wrong.

I have used momentum to determine the velocity of the arrow to be 57.6m/s.

Thank you for the help.

 

[PeroK]

I have used momentum to determine the velocity of the arrow to be 57.6m/s.

Thank you for the help.

That's what I get!

 

[haruspex]

the kinetic energy of the arrow before impact equal the work required to penetrate the arrow (F . x1) & move the block (F . x2).

That would work, but you do not know what work was required to penetrate the block. In your first attempt, you seem to have assumed the force resisting penetration is the same as the frictional force from the table. Clearly that is not the case since we could make that frictional force infinite and still have the arrow penetrate the block.

 

[haruspex]

PS The book answer looks wrong to me. Note first, that the mass of the system is irrelevant:
$$F_f = \mu Mg \ \Rightarrow \ a = -\frac {F_f}{M} = -\mu g = -0.3 \times 9.8 m/s^2 = -2.94 m/s^2$$Now, suppose the book answer is correct and the initial velocity of the block is about $1m/s$. The motion will last about $0.34s$, at an average speed of $0.5m/s$, which gives a distance of about $0.17m$.

My conclusion is that whoever did the book solution did the following erroneous calculation:
$$v = 2\sqrt{as}$$You might need a new book as well!

A difficulty is how to interpret "the velocity of the target immediately after impact".
To get 0.711 m/s you must be using the 0.086mm as the distance traveled "after impact", but that is the distance the block moves from the first contact between the arrow and the block. If that is what "after impact" means then the velocity of the block at that time is zero. It is not going to immediately take off at max speed, yet the arrow continues to penetrate it for a while.

That said, I have not found a way to get the book answer.

Edit:
During the penetration phase, let subscripts A, B denote the arrow and block respectively. If the force required to push the arrow into a fixed block is F then I get the power equation
$m_Av_A\dot v_A+m_Bv_B\dot v_B=F(v_B-v_A)$
And the momentum equation
$m_Av_A+m_Bv_B=m_Au$
where u is the initial speed of the arrow.
From these I get an ODE of the form
$yy'+ay+by'+c=0$.
And there I am stuck.

Perhaps you should give up on that site.

For the problem addressed there, the energy method is appropriate.

 

[PeroK]

A difficulty is how to interpret "the velocity of the target immediately after impact".
To get 0.711 m/s you must be using the 0.086mm as the distance traveled "after impact", but that is the distance the block moves from the first contact between the arrow and the block. If that is what "after impact" means then the velocity of the block at that time is zero. It is not going to immediately take off at max speed, yet the arrow continues to penetrate it for a while.

If the block moves before we consider the collision ended, then the remaining distance is less than $86mm$ and the initial velocity must be less than the $0.711m/s$. It can't be greater.

Moreover, if we assume that the arrow has an initial speed of at least $50m/s$, the its average speed during the collision will be about $25m/s$, and it will only take about $0.002s$ to travel $50mm$. Meanwhile, the accelerating block will travel only about $1mm$ in that time.

That rough calculation shows that the collision is effectively instantaneous.