Problem with bullet collision exercise

[DDarthVader]

Homework Statement

Hello! I'm trying to solve this exercise:
A bullet with initial velocity v and mass m pass through a block of mass M in a pendulum of length L and leave the block with velocity v/2. Which value of v will make the pendulum describe a circumference?
As you can see, English is not my native language.

Homework Equations

The Attempt at a Solution

I think this is a inelastic collision so this is what I did.
$K_{1i}+K_{2i}=K_{1f}+K_{2f}$ but the pendulum has velocity 0 when the bullet hit it so $K_{1i}+0=K_{1f}+K_{2f}$.

Using the information above. $\frac{mv^2}{2}=\frac{mv^2}{8}+\frac{Mv^2}{2}$. I did this because the kinetic energy is conserved so I think I can use this to get to the mechanical energy which will give me $Mg2L=K$. I think this is the way but honestly I don't know how to get the K to make it equal to the potential energy.

 

[Villyer]

Which value of v will make the pendulum describe a circumference?

What is meant by 'describe a circumference'?

 

[DDarthVader]

The bullet will hit the block and make it's center of mass describe a circumference. But it will only happen if the velocity v has a certain value or bigger than this value. I want to know what value is this. I can post a picture or post the .pdf link so you can see the situation.

 

[Villyer]

Again you mention 'describe a circumference', but I don't get what that refers to.

Move in a complete circle?

 

[DDarthVader]

Exactly, it moves in a complete circle.It's path is a complete circle.

 

[gneill]

The bullet passes through the block, so what can you say about the type of collision involved? What is conserved in that sort of collision?

 

[DDarthVader]

The bullet passes through the block, so what can you say about the type of collision involved? What is conserved in that sort of collision?

I think it's a inelastic collision. If I'm correct them the kinetic energy and linear momentum are both conserved. I think I can solve this exercise using the conservation of energy of the system after the collision but I'm not sure how to do this.

PS: Is this situation possible? I mean, I can't see it happening in my mind.

 

[Villyer]

Assuming that the collision is inelastic is a gamble, but momentum is always conserved and is a more solid base for this problem.

 

[gneill]

I think it's a inelastic collision. If I'm correct them the kinetic energy and linear momentum are both conserved. I think I can solve this exercise using the conservation of energy of the system after the collision but I'm not sure how to do this.

PS: Is this situation possible? I mean, I can't see it happening in my mind.

It's a form of inelastic collision where one of the bodies does not remain "attached". The only thing you can count on in an inelastic collision is that momentum will be conserved. Momentum is ALWAYS conserved.

Why don't you consider the momentum of the bullet pre-collision and post-collision? Where do you think the difference will end up?

 

[DDarthVader]

Assuming that the collision is inelastic is a gamble, but momentum is always conserved and is a more solid base for this problem.

In my mind the collision was inelastic because there was no external forces. But there are things like heat and sound so it's not a inelastic collision.

It's a form of inelastic collision where one of the bodies does not remain "attached". The only thing you can count on in an inelastic collision is that momentum will be conserved. Momentum is ALWAYS conserved.

Why don't you consider the momentum of the bullet pre-collision and post-collision? Where do you think the difference will end up?

I tried two things now. First way: by conservation of momentum: $P_{1i}+P_{2i}=P_{1f}+P_{2f}$ but we know that $P_{2i}=0$. Then $P_{1i}=P_{1f}+P_{2f}$.
Using the data the exercise gave to us $mv=m\frac{v}{2}+Mv_{2f}$.

Second way: by conversation of total mechanical energy:
$\frac{1}{2}mv^2=Mg2L$ solving for v I got $v=2 \sqrt{\frac{MgL}{m}}$. Wrong again. What's wrong with this equation? Because in this situation the mechanical energy will be conserved right? So the kinetic energy of the bullet is equal to the potential energy of the block.

 

[Villyer]

Your second way is incorrect because some initial kinetic energy is lost to heat and sound.

 

[DDarthVader]

Your second way is incorrect because some initial kinetic energy is lost to heat and sound.

Ok, I Agree and understand why but if this is incorrect I can't get the right answer. The conservation of momentum won't give me a relation between the velocity v and L.
By the first way I got this answer $v=\frac{2Mv_{2f}}{m}$ and according to the textbook that's wrong.

 

[gneill]

First find an expression for the initial velocity of the block as a function of the initial velocity of the bullet, v. Then consider conservation of energy for the motion of the block as it moves around its trajectory.

 

[DDarthVader]

First find an expression for the initial velocity of the block as a function of the initial velocity of the bullet, v. Then consider conservation of energy for the motion of the block as it moves around its trajectory.

Using conversation of momentum: $P_{1i}+0=P_{1f}+P_{2f}$ and from this equation we get $v_{2}(v)=\frac{vm}{2M}$ and $v_{2}$ is the velocity of the block as a function of the initial velocity of the bullet.
Now considering the conversation of energy for the motion of the block:
$K_i + U_i = K_f + U_f$ using the data we have
$\frac{m(v_{2})^2}{2}=Mg2L$
$(\frac{vm}{2M})^2=\frac{Mg4L}{m}$ solving for v
$v=4\frac{M}{m}\sqrt{\frac{MgL}{m}}$
Is this correct??

 

[gneill]

I think you have to consider what velocity the block must have at the apex of its trajectory in order for it to describe a circular motion...
Hint: The velocity (kinetic energy) will not be zero.

 

[DDarthVader]

The only idea I had until now was that at the apex of its trajectory, the block will have velocity $v=\sqrt{gL}$ so its kinetic energy is $K=M\sqrt{gL}$ but when I plugged this on the equation $\frac{m(v_{2})^2}{2}=Mg2L+M\sqrt{gL}$ ... well guess what... wrong again. *Sigh*

 

[gneill]

The KE at the apex is the initial KE MINUS the PE due to its gain in altitude. Also, KE is $\frac{1}{2}M V^2$, not $MV$.

 

[DDarthVader]

The KE at the apex is the initial KE MINUS the PE due to its gain in altitude. Also, KE is $\frac{1}{2}M V^2$, not $MV$.

Oh, my bad. So you are saying that $K_f$ is $0-2MgL$? Because the initial kinetic energy of the block is zero since its velocity was zero before the bullet hit it. And PE at the apex is $2MgL$.

 

[gneill]

Oh, my bad. So you are saying that $K_f$ is $0-2MgL$? Because the initial kinetic energy of the block is zero since its velocity was zero before the bullet hit it. And PE at the apex is $2MgL$.

No, the "initial" KE of the block in this case is its KE immediately after the "collision" with the bullet.

There are two distinct phases of the problem: The first is the analysis of the bullet-block collision wherein momentum is transferred to the block. The second is the resulting motion of the block given the "kick" it received from the bullet. The "initial" KE for this second phase is the KE the block has according to the momentum that resulted from the collision. It's this "initial" KE that you need to work with for the continuing motion of the block.

 

[DDarthVader]

Ok! Now I'm trying this:
$K_i + U_i=K_f + U_f$ and using what you told me (KE at the apex is KE initial minus PE)
$\frac{m(v_2)^2}{2}=(K_i - U_f)+U_f$ doing the algebra
$\frac{v^2 m^3}{4M}=\frac{v^2m^3-8M^2L}{4M}-Mg2L$
I understand the situation now but still I can't solve this exercise...

 

[azizlwl]

It's a form of inelastic collision where one of the bodies does not remain "attached". The only thing you can count on in an inelastic collision is that momentum will be conserved. Momentum is ALWAYS conserved.

Why don't you consider the momentum of the bullet pre-collision and post-collision? Where do you think the difference will end up?

Energy is conserved but some are transformed to other form other than kinetic energy.

mv₁=Mv_{f_M}+mv_{f_m}
v_{f_M}=(m/2M)v₁

Mg2L=0.5M(v_{f_M})²

 

[DDarthVader]

Energy is conserved but some are transformed to other form other than kinetic energy.

mv₁=Mv_{f_M}+mv_{f_m}
v_{f_M}=(m/2M)v₁

Mg2L=0.5M(v_{f_M})²

Basically you are saying that $K_i+U_i=K_f+U_f \Rightarrow \frac{M(v_{fM})^2}{2}+0=0+Mg2L$ where $v_{fM}=\frac{vm}{2M}$. And that means that $K_f$=0 and gneill was telling me that this true. Or am I getting everything wrong?

 

[gneill]

Ok! Now I'm trying this:
$K_i + U_i=K_f + U_f$ and using what you told me (KE at the apex is KE initial minus PE)
$\frac{m(v_2)^2}{2}=(K_i - U_f)+U_f$ doing the algebra
$\frac{v^2 m^3}{4M}=\frac{v^2m^3-8M^2L}{4M}-Mg2L$
I understand the situation now but still I can't solve this exercise...

You seem to be mixing up the masses. After the collision the only mass you're concerned about is M. If the "initial" velocity of the M is V = (m/M)v/2, then the "initial" KE is (1/2)M[(m/M)v/2]². You shouldn't have any m³ terms in there.

 

[DDarthVader]

Oh, algebra... Well I finally got it! I now understand the math and understand the situation and the exercise is solved! Thank you guys very much and a special thanks to gneill for you patience! :) Now I can finally sleep.Thanksss!

 

[gneill]

Energy is conserved but some are transformed to other form other than kinetic energy.

mv₁=Mv_{f_M}+mv_{f_m}
v_{f_M}=(m/2M)v₁

Mg2L=0.5M(v_{f_M})²

No, this is not correct. What you are implying is that the initial KE of the mass M is just sufficient to take it to the apex, at which point its KE would be zero having traded all its KE for gravitational PE. It would then fall straight down and not hold to a circular trajectory.

 

[gneill]

Oh, algebra... Well I finally got it! I now understand the math and understand the situation and the exercise is solved! Thank you guys very much and a special thanks to gneill for you patience! :) Now I can finally sleep.Thanksss!

No problem. Glad to help!

 

[azizlwl]

No, this is not correct. What you are implying is that the initial KE of the mass M is just sufficient to take it to the apex, at which point its KE would be zero having traded all its KE for gravitational PE. It would then fall straight down and not hold to a circular trajectory.

Thanks error there.
Should be centripetal force(Mv²/L) at apex equal to Mg.