Block and Bullet Collision: Finding the Height of Block B | Homework Problem

[Tanya Sharma]

Homework Statement

A block A of mass M_A = 1 kg is kept on a smooth horizontal surface and attached by a light thread to another block B of mass M_B=2 kg .Block B is resting on ground and thread and pulley are massless and frictionless.A bullet of mass m 0.25 kg moving horizontally with velocity of u=200m/s penetrates through the block and comes out with a velocity of 100m/s.Find the height through which the block B will rise.

Ans :5.21m

Homework Equations

The Attempt at a Solution

When bullet penetrates through block A ,an impulsive force P acts on it.

∫Pdt = 0.25(200-100) = 25Ns

Considering block A,let the impulsive tension be T

∫Pdt - ∫Tdt = 1(v_A)

Considering block B ,

∫Tdt = 2v_B

I guess I am missing something ,is it v_A = v_B because of inextensibility of the string connecting the masses? But that doesn't give correct answer .

I would be grateful if somebody could help me with the problem.

 

[ehild]

The speed of both blocks must be equal. You can assume that the bullet penetrates through the block so fast that you can ignore any displacements of the blocks during that time. But you know their speed from conservation of momentum. And then it is a decelerating motion...

ehild

 

[Tanya Sharma]

Thanks ehild !

Does that mean first an impulsive tension acts which gives the blocks an initial velocity and afterwards a constant tension exists in the string?

The speed of both blocks must be equal.

Due to impulsive tension why can't the initial velocity of block B greater than that of block A i.e v_B > v_A ? In that case block B will be under free fall .

But you know their speed from conservation of momentum.

Conservation of momentum or Impulse momentum theorem ?

 

[ehild]

There can be different approximations. I assumed that both blocks make a system that gets common speed from the bullet. I do not understand your impulsive tensions.

ehild

 

[Saitama]

The speed of both blocks must be equal. You can assume that the bullet penetrates through the block so fast that you can ignore any displacements of the blocks during that time. But you know their speed from conservation of momentum. And then it is a decelerating motion...

ehild

I am not sure but I tried this on my first look at the problem and this approach didn't give me the right answer.

From conservation of momentum, I got speed of $M_A$ equal to $25\,m/s$. Hence, the speed of $M_B$ is $25\,m/s$.

I got the magnitude of deacceleration equal to $20g/3$. Hence, max height achieved by B is:
$$h=\frac{v^2}{2a}=\frac{625\times 3}{40g} \approx 4.6875\,m$$

 

[Tanya Sharma]

How does the bullet give the blocks a common speed ? The bullet passes through the block A in a very short period ,say dt .A large contact force arises between the bullet and the block A .This in turn gives rise to large tension in the string for a short duration .This tension is responsible for the initial velocity of block B .

If it were not for impulsive tension , how else would block B get initial velocity ?

 

[Tanya Sharma]

From conservation of momentum, I got speed of $M_A$ equal to $25\,m/s$. Hence, the speed of $M_B$ is $25\,m/s$.

Momentum is not conserved in this problem . Please refer my attempt in the OP .From that you will get v = 25/3 m/s .

 

[SammyS]

How does the bullet give the blocks a common speed ?
...

It must be assumed that the thread is inextensible . It doesn't stretch !

 

[Tanya Sharma]

It must be assumed that the thread is inextensible . It doesn't stretch !

Okay...But what about the possibility of block B getting a higher initial velocity then block A due to the impulse ?

 

[FermiAged]

Could conservation of energy play a role in the problem?

 

[ehild]

Okay...But what about the possibility of block B getting a higher initial velocity then block A due to the impulse ?

How? that would require a varying tension in the string.

ehild

 

[FermiAged]

Okay...But what about the possibility of block B getting a higher initial velocity then block A due to the impulse ?

That would require block B accelerating relative to block A which implies a force. Other than gravity and the string, is there another force present?

 

[haruspex]

Okay...But what about the possibility of block B getting a higher initial velocity then block A due to the impulse ?

That could indeed happen if there were any elasticity in the string. But how to prove it doesn't when there's no elasticity is an interesting question. I don't find the answers so far convincing.

How? that would require a varying tension in the string

No, only that the impulsive tension is greater than it needs to be to get the blocks moving at the same speed.

That would require block B accelerating relative to block A which implies a force. Other than gravity and the string, is there another force present?

It doesn't require an ongoing force, merely a sufficiently large initial impulsive tension.

 

[SammyS]

Homework Statement

A block A of mass M_A = 1 kg is kept on a smooth horizontal surface and attached by a light thread to another block B of mass M_B=2 kg .Block B is resting on ground and thread and pulley are massless and frictionless.A bullet of mass m 0.25 kg moving horizontally with velocity of u=200m/s penetrates through the block and comes out with a velocity of 100m/s.Find the height through which the block B will rise.

Ans :5.21m
...

I treated it pretty much like an inelastic (not perfectly inelastic) collision of a 0.25 kg bullet (That's one MASSIVE bullet) with a 3 kg object, then used conservation of energy with PE based on the 2 kg block.

I came up with 5.31 m .

 

[nasu]

They assume that the bullet collides with the system of the two blocks, as a whole.
I mean, this way you get their result.

If you look at a more realistic scenarios, it is not even necessary that the bloc B rise from the ground. If block A is long and "soft", so that the bullet looses momentum slowly, the tension in the spring may be always less than the weight of B. At least in principle.

Maybe we can solve a scenario assuming a constant force between bullet and bloc A and an elastic string, to study more interesting/realistic versions of the problem.

Edit. Sammy already posted along the lines of my first paragraph. I've got 5.21 m though.

 

[Tanya Sharma]

That could indeed happen if there were any elasticity in the string.

Since the string is inelastic is it safe to assume that v_B≤v_A ?

No, only that the impulsive tension is greater than it needs to be to get the blocks moving at the same speed.

It doesn't require an ongoing force, merely a sufficiently large initial impulsive tension.

This is exactly the point that makes me unsure that the two blocks move together with the same speed .

Do you agree that initially an impulsive tension acts and thereafter a constant tension exists in the string ?

If you look at my attempt in the OP ,if value of ∫Tdt is quite large than the block B might have a velocity larger than that of block A.

I still do not understand how can we assume that the two blocks move with the same speed.

This concept of impulsive tension comes up quite often in the problems ,so I would like to be sure about it.

 

[Tanya Sharma]

I treated it pretty much like an inelastic (not perfectly inelastic) collision of a 0.25 kg bullet (That's one MASSIVE bullet) with a 3 kg object, then used conservation of energy with PE based on the 2 kg block.

I came up with 5.31 m .

How did you assume that the two blocks move together ? If the initial velocity of block A is more than that block B ,the inextensibility of the string makes sure that the two move together .But the inextensibility of the string doesn't rule out that the velocity of B can't be more than that of A.

 

[SammyS]

How did you assume that the two blocks move together ? If the initial velocity of block A is more than that block B ,the inextensibility of the string makes sure that the two move together .But the inextensibility of the string doesn't rule out that the velocity of B can't be more than that of A.

What mechanism is present in this problem which can make the speed of Block B be more than that of Block A ?

 

[Tanya Sharma]

What mechanism is present in this problem which can make the speed of Block B be more than that of Block A ?

Impulsive tension

 

[ehild]

What do you call "impulsive tension"? I learned about momentum, I know that the force is equal to the time derivative of momentum, and the momentum of a closed system is conserved. I also know that conservation of momentum does not apply if there is some "impulsive" external force, from a constraint, that can become indefinitely large during the interaction. Is there any constraint for the motion of the bodies along the string? Both are free to move.

There is no such thing that infinite force in real life. And inextensible strings do not exist. All strings are extensible a bit. If you think that the connected bodies do not move together, you should take the elasticity of the string into consideration. It is clear that the bullet interacts with block A and it starts to move. That motion causes stress in the string, it stretches a bit. That disturbance travels along the string and reaches block B, and causes to move it, when the tension at that place overcomes gravity. .The acceleration of B is the effect, and the initial motion of the first block is the cause. It can happen, that B moves faster than A when the bullet leaves A, but you need to derive it from the properties of the string. The problem says that the string is inextensible, that means the blocks start to move together, as parts of a rigid body. There is also the possibility, that you push block B upward, and this way it moves faster than block A. That makes the string slack. No force opposes gravity at B, and no retarding tension acts to block A. While it interacts with the bullet, it will accelerate faster than in case of a taut string. After the bullet leaves, it will move with a constant speed, while block B will decelerate because of gravity. Sooner or later the string becomes taut and the blocks get to move together.

As the bullet and the blocks and the string make a system, the component of the momentum along the string is conserved as the only external force is gravity and it is constant, so its effect can be ignored during the interaction with the bullet. You do not know what happens exactly during the interaction with the bullet and after a short time, but the result of the collision process is a taut string and two blocks moving together.

ehild

 

[Tanya Sharma]

In the attachment ,there are two blocks A and B attached by a light inextensible string lying on a frictionless table .A sharp impulse ∫Fdt = 100Ns is given to A .As a result a large tension develops in the string.

Let the tension in the string be T .

Considering block A ,

∫Fdt-∫Tdt = 10v_A

or, 100 - ∫Tdt = 10v_A

Considering block B ,

∫Tdt = 2v_B

Now ,suppose ∫Tdt = 20Ns

In that case v_A = 8m/s and v_B = 10m/s .The two blocks don't move together despite inextensible string.

Please point out the flaw in the set up . May be I am overlooking certain assumptions .

Thanks

 

[ehild]

From where do you know ∫Tdt? Do you have any reason why it should be 20 Ns, or anything else?

We use conservation of momentum for closed systems, as we do not know the details of interaction. We know that the system of the two blocks got 100 Ns impulse. That means, the change of momentum is 100 kgm/s. You do not know anything about the time dependence of tension in the string.

ehild

 

[haruspex]

As mentioned, all strings are extensible in practice. i suggest the way to make progress here is to suppose an elastic string, solve it, then see what happens as the string's modulus tends to infinity.

 

[Tanya Sharma]

From where do you know ∫Tdt? Do you have any reason why it should be 20 Ns, or anything else?

It is just an assumption .I do not have any reason why it can not be 20Ns .Do you have a reason why it cannot be 20Ns ? Do you think it is an impossible value ?

This is just an example to illustrate situation similar to OP.

We use conservation of momentum for closed systems, as we do not know the details of interaction. We know that the system of the two blocks got 100 Ns impulse. That means, the change of momentum is 100 kgm/s. You do not know anything about the time dependence of tension in the string.

ehild

Agreed...But we do not know how that momentum is distributed .This does not explain why the two blocks must move together .

I am not trying to convey that the blocks will not or cannot move with the same speed .All I am saying is that the two blocks may or may not move together .If the impulse due to tension is sufficiently small they may move together .But if the impulse is large block B may have larger speed than block A and the two do not move together .

Is it an assumption in intro physics problems that blocks connected with string surely must move together ?

 

[ehild]

It is just an assumption .I do not have any reason why it can not be 20Ns .Do you have a reason why it cannot be 20Ns ? Do you think it is an impossible value ?

This is just an example to illustrate situation similar to OP.

Well, you can assume that it is -10000 Ns as well. And what is it good for, to make wild guesses? Do you think Physics works this way?

Agreed...But we do not know how that momentum is distributed .This does not explain why the two blocks must move together .

I am not trying to convey that the blocks will not or cannot move with the same speed .All I am saying is that the two blocks may or may not move together .If the impulse due to tension is sufficiently small they may move together .But if the impulse is large block B may have larger speed than block A and the two do not move together .

Is it an assumption in intro physics problems that blocks connected with string surely must move together ?

If the string is always taut and inextensible, yes, the masses move together after the impulsive force does not act any more.

Try to solve the problem with a spring instead of string. Assume a massless spring, and Hook's Law valid. Assume constant force F for a very short time Δt. Both masses in rest initially. You can solve this problem.
You will get that A will move and B stays in rest at the beginning of the impulsive action. There will be an oscillatory term in the motion. But this oscillation ceases sooner or later as there is always some internal friction in the spring.

ehild

 

[nasu]

In the attachment ,there are two blocks A and B attached by a light inextensible string lying on a frictionless table .A sharp impulse ∫Fdt = 100Ns is given to A .As a result a large tension develops in the string.

Let the tension in the string be T .

Considering block A ,

∫Fdt-∫Tdt = 10v_A

or, 100 - ∫Tdt = 10v_A

Considering block B ,

∫Tdt = 2v_B

Now ,suppose ∫Tdt = 20Ns

In that case v_A = 8m/s and v_B = 10m/s .The two blocks don't move together despite inextensible string.

Please point out the flaw in the set up . May be I am overlooking certain assumptions .

Thanks

For inextensible string, v_B cannot be less than v_A.
The inextensble string allows for v_B > v_A but then the tension is zero.

So as long as there is tension in the string, the two speed must be equal. If we start with zero initial velocity of B, the tension accelerates B, so we must have some non-zero tension at least for a while. So that means they have the same speed for a while and then they must have the same acceleration (otherwise they will not have same speed over a finite interval). And if they have the same acceleration...

So again, if you want to get a more realistic image (and understanding), you need to drop "inextensible", as it was already suggested several times.
Then you can have different speeds and nonzero tension.