ch 7.3 help

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ch 7.3 help

Post  guest07 on Tue Oct 14, 2008 4:17 pm

anyone have any clue?

guest07
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Re: ch 7.3 help

Post  Guesto on Tue Oct 14, 2008 5:26 pm

Question 2:
M1V1i+M2V2=M1V1f+M2V2f
Masses are equal, so:
M(V1i+V2i)=M(V1f+V2f)
Initial velocity of object 2 is zero, and because they are of equal mass, the final velocity of object 1 will be zero (think about it):
M(V1i)=M(V2)
Divide out the masses and find that... V1=V2.
Simply put, because the masses are equal, all of the velocity is transferred from the incident object to the target object. PEACE.

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Re: ch 7.3 help

Post  Guesto on Tue Oct 14, 2008 5:29 pm

Question 4:
They give you the ratios of the velocities as V:
1/V gives you the ratio of the masses. Don't really know why.
So those are the easy ones anyone get the 1st or 3rd one?

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Number one

Post  super Mo on Thu Oct 16, 2008 3:11 pm

Use the formula below and solve for h:
1/2 (m/m+M^2) * v^2=gh
h = (1/2 (m/m+M)^2 * v^2)/g

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Re: ch 7.3 help

Post  blah on Thu Oct 16, 2008 3:42 pm

Number two keeps getting marked wrong on maple TA for me. Is it exactly the same as the velocity given, or what equations do you use to determine it?

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question 2

Post  hwilson on Thu Oct 16, 2008 4:21 pm

I was getting it wrong too u have to add negative... So if velocity is 14.55 final ans is -14.55.

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anyone got numbers 1, 2, 3, 4, with the correct signs?

Post  spetro71 on Thu Oct 16, 2008 4:49 pm

i really need help with these

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Q#3

Post  p on Thu Oct 16, 2008 5:14 pm

Q#3
An object makes an elastic head-on collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.5432 what is the velocity of the incident object after the collision in multiples of its incident velocity (see sheet 25,29)? Give the result with the appropriate sign taking the incident velocity as positive.

anybody has any idea how to do this question. thanks

p
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Re: ch 7.3 help

Post  guestttt on Thu Oct 16, 2008 7:23 pm

1, 2, 3 are all negative answers
4 is positive

However, I keep getting 4 wrong. any suggestions?

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need help with # 3 and 4

Post  spetro71 on Thu Oct 16, 2008 8:35 pm

anyone got any ideas

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#3/4

Post  lanthony on Thu Oct 16, 2008 9:27 pm

I have tried everythign with these two problems. Can anyone offer any direction (that 1 over V thing is not working at all).

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Question 3

Post  student on Thu Oct 16, 2008 10:44 pm

Here's what I did
An object makes an elastic head-on collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.7014 what is the velocity of the incident object after the collision in multiples of its incident velocity (see sheet 25,29) ? Give the result with the appropriate sign taking the incident velocity as positive.
You know the ratio is 0.7014, to keep things simple I said m1=0.7014 and m2= 1 since dividing m1/m2= 0.7014
v1=(m1-m2)/m1+m2)
v1=(0.7014-1)/1+0.7014)= -0.1755

Has anyone gotten 4?

student
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Question 4 Possible Solution

Post  S on Fri Oct 17, 2008 12:03 am

For number 4, did anyone just try using the value of the incident velocity. The reason I say this is because V1= (m2/m1)V2 +V1'....

I can't do it because my maple ta is down, but i'd love to know. Smile

S
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question 2

Post  stuckk on Fri Oct 17, 2008 12:19 am

Does anyone know how to solve number two?
I've tried but I can't seem to get it....

Thank you!

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Re: ch 7.3 help

Post  guest3 on Fri Oct 17, 2008 12:20 am

For number 4 I used this:

2*(v2'/v1) - 1 = m1/m2

Seems to work.

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Q 2

Post  student on Fri Oct 17, 2008 4:50 pm

use the equation

0.5((bulletmass/(bulletmass+woodmass))^2*velocity^2=g*h

divide the bullet mass by the bullet mass plus the wood's mass, square it then multiply by 1/2. square the velocity and multiply that by your current ratio. Then divide by 9.81 to get your height.

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