State whether the distance x for the block would be greater/less/same

[paulimerci]

Homework Statement

Attached problem below!

Relevant Equations

Conservation of momentum and Conservation of energy

I was able to solve for a, b, c, and d.
I don't know whether I'm saying it right for part e.
The first scenario is a perfectly inelastic collision. After impact, the bullet transfers momentum to the block, causing the block to move (it was initially at rest), and the bullet + block fall at a distance of x. However, when we perform energy conservation for the block-bullet system, the final velocity is less than the initial velocity of the block.
The second scenario is a perfectly elastic collision. During the impact, the bullet transferred some of its momentum to the block and leaves with velocity less compared to its initial velocity but a higher velocity compared to the block, and thereby the block should have landed at a distance less than x.
Was it right?

 

[haruspex]

the final velocity is less than the initial velocity of the block.

True, but it asks about energy, not velocity.

The second scenario is a perfectly elastic collision.

No.

a higher velocity compared to the block, and thereby the block should have landed at a distance less than x.

Yes, though you could fill in one more step in that logic.

 

[paulimerci]

True, but it asks about energy, not velocity.

No.

Yes, though you could fill in one more step in that logic.

The second scenario is an elastic collision, the bullet doesn’t stick to the block. The collision was inelastic to elastic after impact. what type of collision should I write that as?

 

[paulimerci]

True, but it asks about energy, not velocity.

No.

Yes, though you could fill in one more step in that logic.

Guide me what step should I add?

 

[haruspex]

The second scenario is an elastic collision, the bullet doesn’t stick to the block.

It doesn’t bounce off either. It is neither perfectly elastic nor perfectly inelastic.
The standard terminology (which I dislike) uses "elastic" to mean perfectly elastic. Anything else is inelastic. Coalescence, as in the first part, is perfectly inelastic.

Guide me what step should I add?

You wrote that the block's velocity in the second scenario will be less than that of the bullet in the second scenario. To show the distance is less in the second scenario than in the first, you need to refer to something that is different between the two scenarios.

 

[paulimerci]

It doesn’t bounce off either. It is neither perfectly elastic nor perfectly inelastic.
The standard terminology (which I dislike) uses "elastic" to mean perfectly elastic. Anything else is inelastic. Coalescence, as in the first part, is perfectly inelastic.

You wrote that the block's velocity in the second scenario will be less than that of the bullet in the second scenario. To show the distance is less in the second scenario than in the first, you need to refer to something that is different between the two scenarios.

Okay, As the block leaves the table it takes a projectile motion. Do I have to use kinematic equations?

 

[haruspex]

n. Do I have to use kinematic equations?

No, just indicate what is different between the two scenarios as the block leaves the table, and why that difference occurs.

 

[kuruman]

The second scenario is a perfectly elastic collision.

I don't see it that way. When you have a perfectly elastic collision, the final velocities of the colliding objects are strictly determined from the initial velocities. Specifically in this case, since the block is 100 times more massive than the bullet, the bullet in a perfectly elastic collision should bounce back. That is one solution implied by the two conservation equations. The other solution that conserves energy and momentum is for the bullet to go through unmolested and is the solution that is usually ignored because it is just an affirmation of Newton's first law. Here, bullet exchanges energy and momentum with the block, as it goes through the block which fits neither of the two possible outcomes.

 

[paulimerci]

I don't see it that way. When you have a perfectly elastic collision, the final velocities of the colliding objects are strictly determined from the initial velocities. Specifically in this case, since the block is 100 times more massive than the bullet, the bullet in a perfectly elastic collision should bounce back. That is one solution implied by the two conservation equations. The other solution that conserves energy and momentum is for the bullet to go through unmolested and is the solution that is usually ignored because it is just an affirmation of Newton's first law. Here, bullet exchanges energy and momentum with the block, as it goes through the block which fits neither of the two possible outcomes.

Okay, I agree with what you said. Thank you @kuruman and @haruspex for taking time to read this and explain.

 

[paulimerci]

No, just indicate what is different between the two scenarios as the block leaves the table, and why that difference occurs.

That is what I posted in #1. I need to correct the error I made for scenario two. You mean that?

 

[haruspex]

That is what I posted in #1.

I cannot see that you did.
Just concentrate on the point, within each scenario, where the block is about to leave the table. What is different for the block in the second scenario from its attributes in the first?

 

[paulimerci]

Since the block is 100 times larger than the mass of the bullet, it should have eventually bounced back, which didn't happen in both scenarios. In the first scenario, it's a perfectly inelastic collision, and in the second scenario, it's neither elastic nor inelastic.
In the first scenario, the force of the bullet was much greater, which should have eventually reduced as it collided and stuck with the block, which means its impact time is greater and the force of collision is less (similar to air bags? I'm not sure). In the second scenario, the impact time is less and the force of the collision is greater. And that's the reason the block landed at a distance less than x. I'm I right?

 

[haruspex]

Since the block is 100 times larger than the mass of the bullet, it should have eventually bounced back, which didn't happen in both scenarios

Irrelevant.

in the second scenario, it's neither elastic nor inelastic.

No, as I wrote, it is inelastic, just not perfectly elastic.

In the first scenario, the force of the bullet was much greater

What do you mean by "the force of the bullet"? The force it exerts on the block? You cannot be sure of that, and it is not necessary to be. Perhaps the first block was longer, though the same mass.

which means its impact time is greater

Unlikely.

and the force of collision is less

You just said it was much greater.

In the second scenario, the impact time is less and the force of the collision is greater. And that's the reason the block landed at a distance less than x.

I don't see the logic there.

It's not complicated. Just try to answer the question I posed in post #11. At the point where the block is about to leave the table, what determines how far it will go before landing?

 

[paulimerci]

Irrelevant.

No, as I wrote, it is inelastic, just not perfectly elastic.

What do you mean by "the force of the bullet"? The force it exerts on the block? You cannot be sure of that, and it is not necessary to be. Perhaps the first block was longer, though the same mass.

Unlikely.

You just said it was much greater.

I don't see the logic there.

It's not complicated. Just try to answer the question I posed in post #11. At the point where the block is about to leave the table, what determines how far it will go before landing?

Okay, the horizontal component of the velocity is different in both scenarios.

 

[haruspex]

Okay, the horizontal component of the velocity is different in both scenarios.

Right. So step back a bit further. Why is the horizontal velocity of the block different?

 

[paulimerci]

Right. So step back a bit further. Why is the horizontal velocity of the block different?

because of the transfer of energy from the bullet to the block.

 

[haruspex]

because of the transfer of energy from the bullet to the block.

This is an inelastic collision. What can you be more precise about than transfer of energy?

 

[paulimerci]

In an inelastic collision, kinetic energy is not conserved, but momentum is.

 

[haruspex]

In an inelastic collision, kinetic energy is not conserved, but momentum is.

Right. So can you use conservation of momentum to show that the block will acquire less velocity if the bullet passes through it?

 

[paulimerci]

Right. So can you use conservation of momentum to show that the block will acquire less velocity if the bullet passes through it?

Okay,
Applying Conservation of momentum,
$$p_i = p_f$$
$${m_b v_b}_i + {m_w v_w}_i = {m_b v_b}_f + {m_wv_w}_f$$
$$ {v_w}_f = \frac{{m_b v_b}_i - {m_b v_b}_f}{m_w} $$

The final velocity of the bullet is not given.

 

[haruspex]

Okay,
Applying Conservation of momentum,
$$p_i = p_f$$
$${m_b v_b}_i + {m_w v_w}_i = {m_b v_b}_f + {m_wv_w}_f$$
$$ {v_w}_f = \frac{{m_b v_b}_i - {m_b v_b}_f}{m_w} $$

The final velocity of the bullet is not given.

Right, but you know that in the first scenario $v_{bf}=v_{wf}$ and in the second scenario $v_{bf}>v_{wf}$. If in your equation above you increase $v_{bf}$ what happens to $v_{wf}$?

 

[paulimerci]

Right, but you know that in the first scenario $v_{bf}=v_{wf}$ and in the second scenario $v_{bf}>v_{wf}$. If in your equation above you increase $v_{bf}$ what happens to $v_{wf}$?

It($v_wf$) will decrease.

 

[haruspex]

It($v_wf$) will decrease.

So the distance to hitting the floor will ….?

 

[paulimerci]

So the distance to hitting the floor will ….?

decrease too!

 

[haruspex]

decrease too!

Bingo.

 

[paulimerci]

Bingo.

Thank you so much @haruspex for taking time and helping me think and understand the problem. I truly appreciate that!