Can somebody check my answer for momentum

[toesockshoe]

Homework Statement

A ball with mass M moving horizontally at a speed v, collides elastically with a block of mass 3m that is initially hanging at rest from a celing on the end of a wire of length L. Find the maximum angle through which the block swings after it is hit.

Homework Equations

F=dp/dt

The Attempt at a Solution

I split the problem into 2 parts. Part 1 is trying to find the initial velocity of the block and part 2 is trying to find the angle.

PART 1:
let M1 equal the mass of ball and M2 be the mass of the block. (M2=3M1) V1 be the velocity of the ball abd V2 the velcotiy of the block.

$F=\frac{dp}{dt}$

No forces in the x direction:

$0 = \Delta p$
$P_i = P_f$
$M_1V_1=-M_1V_1+M_2V_2$
$V_2=\frac{2M_1V_1}{M_2}$

PART 2:

you can use energy because the contact is elastic... so conservation of energy is valid:

system mass, earth, and bloc

W=deltaE
$0=-\frac{1}{2}mv_2^2+mgh$
we found v2 in Part 1... substitute it and simply:
$h = \frac{2M_1^2V_1^2}{gM_2^2}$
$(L-\frac{2M_1^2V_1^2}{gM_2^2})/L$
$\theta = cos^{-1}((L-\frac{2M_1^2V_1^2}{gM_2^2})/L)$
M2 = 3M1...
$\theta = cos^{-1}((L-\frac{2M_1^2V_1^2}{g3M_1^2})/L)$

is this correct?

 

[nrqed]

Homework Statement

A ball with mass M moving horizontally at a speed v, collides elastically with a block of mass 3m that is initially hanging at rest from a celing on the end of a wire of length L. Find the maximum angle through which the block swings after it is hit.

Homework Equations

F=dp/dt

The Attempt at a Solution

I split the problem into 2 parts. Part 1 is trying to find the initial velocity of the block and part 2 is trying to find the angle.

PART 1:
let M1 equal the mass of ball and M2 be the mass of the block. (M2=3M1) V1 be the velocity of the ball abd V2 the velcotiy of the block.

$F=\frac{dp}{dt}$

No forces in the x direction:

$0 = \Delta p$
$P_i = P_f$
$M_1V_1=-M_1V_1+M_2V_2$

This step is incorrect. If that was true, you would have more kinetic energy at the end then you had at the beginning, which is not possible in a collision. The ball does not recoil at the same speed it was coming in. Elastic means that kinetic energy is conserved, not that ball 1 rebounds at the same speed (a ball rebounds at the same speed in an elastic collision only when it hits a stationary object)

 

[toesockshoe]

This step is incorrect. If that was true, you would have more kinetic energy at the end then you had at the beginning, which is not possible in a collision. The ball does not recoil at the same speed it was coming in. Elastic means that kinetic energy is conserved, not that ball 1 rebounds at the same speed (a ball rebounds at the same speed in an elastic collision only when it hits a stationary object)

alright, i understand. but $V_{12} = V_{i1E}$ correct? V_(12) is the velocity of 1 respect to 2 and V_(i1E) is the initial velocity of 1 respect to the Earth (or ground). my teacher then gave us the following mnemonic:

$V_{1E} = V_{12}+V_{2E}$ where E stands for Earth (or the ground).

then because V_12 is the same as V_i1E, we can say

$M_1V_{i1G} = -M_1V_{f1G}+M_2V_2$
$M_1V_{i1G} = -M_1V_{i1G}-M_1V_{2G} +M_2V_{2G}$ (through the mnemonic i mentioned 3 lines above)
$V_{2G} = \frac{2M_1V_{i1G}}{M_2-M_1}$

is that correct?

 

[toesockshoe]

Bump... Please see comments

 

[theodoros.mihos]

Please, convert all your data to the same reference frame. You cannot write an equation with variables in different frames.

 

[toesockshoe]

Please, convert all your data to the same reference frame. You cannot write an equation with variables in different frames.

arent all my velocity's in reference to the ground?

 

[theodoros.mihos]

Your answer on first post is ok except a 2 factor that must be over.

 

[toesockshoe]

Your answer on first post is ok except a 2 factor that must be over.

i thought my original post was completely incorrect? i said the velocity in reference to the ground were the same before and AFTER the contact... tahts not true... the velocity in reference to M2 that's the same before and after the impact. i made the correction in my comment.

 

[haruspex]

$M_1V_{i1G} = -M_1V_{i1G}-M_1V_{2G} +M_2V_{2G}$
is that correct?

No. How did you lose V_f1G?

i thought my original post was completely incorrect?

It was.
Since it's elastic, you can use conservation of KE. That gets a bit messy because of the quadratics, but combining with conservation of momentum, when the smoke clears, you end up with Newton's Experimental Law for the case R=1, namely, that the relative velocities reverse: V_1i-V_2i=-(V_1f-V_2f). It's well worth remembering this. But note: it only applies for the components of motion perpendicular to the contact plane. No difficulty in the present case, but has to be borne in mind for oblique contacts.

 

[toesockshoe]

No. How did you lose V_f1G?

It was.
Since it's elastic, you can use conservation of KE. That gets a bit messy because of the quadratics, but combining with conservation of momentum, when the smoke clears, you end up with Newton's Experimental Law for the case R=1, namely, that the relative velocities reverse: V_1i-V_2i=-(V_1f-V_2f). It's well worth remembering this. But note: it only applies for the components of motion perpendicular to the contact plane. No difficulty in the present case, but has to be borne in mind for oblique contacts.

ok so using kinetic energy:

lets say the system is the ball and the block... ball as mass1 and block as mass2.
there are no forces in the x direction so:

$W=\Delta E$
$0 = \Delta KE_1 + \Delta KE_2$
$0=\frac{1}{2}m_1v_{i1G}^2-\frac{1}{2}m_1v_{f1G}^2+\frac{1}{2}m_2v_{i2G}^2-\frac{1}{2}m_2v_{f2G}^2$
now how do i bring in the conservation of momentum into this?

 

[haruspex]

ok so using kinetic energy:

lets say the system is the ball and the block... ball as mass1 and block as mass2.
there are no forces in the x direction so:

$W=\Delta E$
$0 = \Delta KE_1 + \Delta KE_2$
$0=\frac{1}{2}mv_{i1G}^2-\frac{1}{2}mv_{f1G}^2+\frac{1}{2}mv_{i2G}^2-\frac{1}{2}mv_{f2G}^2$
now how do i bring in the conservation of momentum into this?

You need to distinguish the two mass variables.
Write out the conservation of momentum equation using the same variables.
Note that you can factorise the two parts of your KE equation.

 

[toesockshoe]

You need to distinguish the two mass variables.
Write out the conservation of momentum equation using the same variables.
Note that you can factorise the two parts of your KE equation.

alright, the conservation of momentum (before and after the collision for m1) using the same variables is:

$m_1v_{i1G}=-m_1v_{f12}$ tell me if this step is correct..
sorry ignore this. i misunderstood. ill come up with anotehr reply in a few minutes after i work it out,

 

[toesockshoe]

You need to distinguish the two mass variables.
Write out the conservation of momentum equation using the same variables.
Note that you can factorise the two parts of your KE equation.

ok so the conservation of momentum gives me the following right?

$0=\Delta p$
$P_i = P_f$
$m_1v_{i1G} = -m_1v_{f1g}+m_2v_{f2g}$
right?

also factoring the KE equation gives me:
$m_1(v_{i1g}^2-v_{f1g}^2)+m_2(v_{i2g}^2-v_{f2g}^2)$ right?

 

[haruspex]

ok so the conservation of momentum gives me the following right?

$0=\Delta p$
$P_i = P_f$
$m_1v_{i1G} = -m_1v_{f1g}+m_2v_{f2g}$
right?

also factoring the KE equation gives me:
$m_1(v_{i1g}^2-v_{f1g}^2)+m_2(v_{i2g}^2-v_{f2g}^2)$ right?

We can drop the g subscripts, yes? And to keep it general for now, introduce v_i2 into the momentum equation, and for clarity, measure all velocities as positive in the same direction:
$m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2}$
$m_1(v_{i1}^2-v_{f1}^2)+m_2(v_{i2}^2-v_{f2}^2)=0$
Rearrange each equation to have m1 on one side and m2 on the other.
You didn't yet factorise the energy equation. How do you factorise $a^2-b^2$?

 

[toesockshoe]

We can drop the g subscripts, yes? And to keep it general for now, introduce v_i2 into the momentum equation, and for clarity, measure all velocities as positive in the same direction:
$m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2}$
$m_1(v_{i1}^2-v_{f1}^2)+m_2(v_{i2}^2-v_{f2}^2)=0$
Rearrange each equation to have m1 on one side and m2 on the other.
You didn't yet factorise the energy equation. How do you factorise $a^2-b^2$?

ok so its:
$m_1(v_{i1}-v{f1})(v_{i1}+v{f1}) + m_2(v_{i2}-v{f2})(v_{i2}+v{f2}) =0$
then
$m_1(v_{i1}-v{f1})(v_{i1}+v{f1}) = -m_2(v_{i2}-v{f2})(v_{i2}+v{f2})$

 

[haruspex]

$m_1(v_{i1}-v_{f1})(v_{i1}+v_{f1}) = -m_2(v_{i2}-v_{f2})(v_{i2}+v_{f2})$

Yes, and rearranging the momentum equation?

 

[toesockshoe]

Yes, and rearranging the momentum equation?

v_{f2}=\frac{m_1(V_{i1}+v_{f1})}{m2} [/itex]
not you can plug in vf2 into the KE equation and vi2 is 0 so you can just throw vi2 away right? then the only unknown you have is vf1 and then you have it solved correct?

 

[haruspex]

v_{f2}=\frac{m_1(V_{i1}+v_{f1})}{m2} [/itex]
not you can plug in vf2 into the KE equation and vi2 is 0 so you can just throw vi2 away right? then the only unknown you have is vf1 and then you have it solved correct?

Sure, but what I'm trying to guide you to is the elastic form of Newton's Experimental Law, which is a very handy equation to know.
You have $m_1(v_{i1}-v_{f1})(v_{i1}+v_{f1}) = -m_2(v_{i2}-v_{f2})(v_{i2}+v_{f2})$ and $m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2}$.
Rewriting the latter as $m_1v_{i1}-m_1v_{f1} = m_2v_{f2}-m_2v_{i2}$, can you see how to combine these two equations to get one that does not involve any masses?

 

[toesockshoe]

Sure, but what I'm trying to guide you to is the elastic form of Newton's Experimental Law, which is a very handy equation to know.
You have $m_1(v_{i1}-v_{f1})(v_{i1}+v_{f1}) = -m_2(v_{i2}-v_{f2})(v_{i2}+v_{f2})$ and $m_1v_{i1}+m_2v_{i2} = m_1v_{f1}+m_2v_{f2}$.
Rewriting the latter as $m_1v_{i1}-m_1v_{f1} = m_2v_{f2}-m_2v_{i2}$, can you see how to combine these two equations to get one that does not involve any masses?

uhm can you divide them?

also vi2 is 0 in this case right?

 

[haruspex]

uhm can you divide them?

also vi2 is 0 in this case right?

Yes, and yes.

 

[toesockshoe]

Yes, and yes.

oh so you get, $v_{i1}+v_{f1} = v_{i2} + v_{f2}$

so can you use this equation for EVERY elastic problem? so when I am doing my homework, i can start of my problems with this forumla and still get right answers? ( i know on the test i would probably havve to start from the beginning).

also, can you respond to me question in: https://www.physicsforums.com/threads/help-on-momentum-problem.816379/

 

[haruspex]

so can you use this equation for EVERY elastic problem?

Yes, and it can be generalised to the partly inelastic case. Search for Newton's Experimental Law or Newton's Law of Restitution.

can you respond to me question in: https://www.physicsforums.com/threads/help-on-momentum-problem.816379/

Apply NEL there too.

 

[toesockshoe]

Yes, and it can be generalised to the partly inelastic case. Search for Newton's Experimental Law or Newton's Law of Restitution.

Apply NEL there too.

alright, i will thanks! but is the way i did it still valid? are there any logical inconsistencies in the way i tried to solve in that far?

 

[haruspex]

alright, i will thanks! but is the way i did it still valid? are there any logical inconsistencies in the way i tried to solve in that far?

It's ok, but as I wrote (here and there) it is better to keep all velocities as positive in the same direction. If the answer you get is a negative v_1f then you will know it bounced back, but it won't always. (Question for you: under what circumstances will it bounce back?)

 

[toesockshoe]

It's ok, but as I wrote (here and there) it is better to keep all velocities as positive in the same direction. If the answer you get is a negative v_1f then you will know it bounced back, but it won't always. (Question for you: under what circumstances will it bounce back?)

hmmm, it would bounce back as long as the other mass is going in the other way... going in the way it was pushed. how can mass1 not come back? if its perfectly inelastic? also, how exactly does perfectly inelastic look? would the two masses just stick and go the same way (the direction in which the first mass pushed the second mass)?

 

[haruspex]

how can mass1 not come back?

Let's fix on the elastic case, m2 being initially at rest.
You have $v_{i1}+v_{f1} = v_{f2}$ and $m_1v_{i1}=m_1v_{f1} +m_2v_{f2}$. We're not interested in $v_{f2}$ now, so combine the equations to eliminate that. Look at what that tells you when $v_{f1} >0, =0, <0$

 

[toesockshoe]

Let's fix on the elastic case, m2 being initially at rest.
You have $v_{i1}+v_{f1} = v_{f2}$ and $m_1v_{i1}=m_1v_{f1} +m_2v_{f2}$. We're not interested in $v_{f2}$ now, so combine the equations to eliminate that. Look at what that tells you when $v_{f1} >0, =0, <0$

i got $v_f=v_1\frac{m_1-m_2}{m_1+m_2}$ all the velocities are of mass 1... if m2 is greater than m1 then the vf is negative right?

 

[haruspex]

i got $v_f=v_1\frac{m_1-m_2}{m_1+m_2}$ all the velocities are of mass 1... if m2 is greater than m1 then the vf is negative right?

Yes!

 

[toesockshoe]

Yes!

wait what the heck? if a ball of mass 2kg hits a ball at rest of 10kg,,,, then both balls will move forward? if this is either elastic or inelastic... (just not completeley inelastic)?

oh nvm, its the other way around... that makes sense.

 

[toesockshoe]

Yes!

also how exactly does perfectly inelastic look? i can't find animations online... if this problem was completely inelastc would the masses stick together or something?

 

[haruspex]

if this problem was completely inelastc would the masses stick together or something?

Yes. They would 'coalesce'.