Homework Help: Pendulum projectile momentum question

1. Feb 4, 2012

jackneedshelp

1. The problem statement, all variables and given/known data
A 0.010 kg pendulum bob is dropped from a height (h) above its equilibrium position. When the bob reaches its equilibrium position, the string breaks and the bob now acts as a projectile. After the string breaks the bob falls 1.5 m while moving 2.0 m horizontally. Calculate the height (h) from which the bob was released.

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2012

tiny-tim

welcome to pf!

hi jackneedshelp! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

3. Feb 4, 2012

jackneedshelp

I don't even know where to start! It should have something to do with momentum and impulse, but I can't even get started on it! Please help!

4. Feb 4, 2012

tiny-tim

first find the speed of the bob (as a function of h) when the string breaks …

show us what you get​

5. Feb 4, 2012

jackneedshelp

I set potential energy equal to kinetic energy, and solved for v.
v = sqrt(19.62h)

6. Feb 4, 2012

tiny-tim

yup!

now use that as the initial horizontal velocity of a projectile …

write standard constant acceleration equations for x and y (separately)

7. Feb 4, 2012

jackneedshelp

so I did the horizontal part, and got the horizontal acceleration to be 4.905h.
how do I do the vertical part?

8. Feb 4, 2012

tiny-tim

how can there be any horizontal acceleration?

there's no horizontal force

9. Feb 4, 2012

jackneedshelp

so I'm solving for time?

10. Feb 4, 2012

jackneedshelp

or final velocity?

11. Feb 4, 2012

tiny-tim

write out some equations!!

12. Feb 4, 2012

jackneedshelp

d = 2.0m
vi = sqrt(19.62h)
vf = 0
t = ?

2 = (0+sqrt(19.62h))/2 *t
t = 4 / (sqrt(19.62h))

is that what I'm supposed to do?

13. Feb 4, 2012

tiny-tim

something like that

but vf isn't 0 (and you aren't told what it is)

and that's the wrong equation anyway

this time, first write out the equation you're using, in letters,

then write it out with the numbers in​

14. Feb 4, 2012

jackneedshelp

vf srd = vi srd + 2ad
vf srd = (sqrt(19.62h))srd + 2(0)(2)
vf = sqrt(19.62h)

so vf = vi

15. Feb 4, 2012

tiny-tim

well, yes, but that's a bit obvious

horizontally, a = 0, so v is constant

you need an equation with t in it ​

16. Feb 4, 2012

jackneedshelp

d = vi t + (0.5)(a)(t)sqrd
2 = sqrt(19.62h) t + 0
t = 2 / sqrt(19.62h)

17. Feb 4, 2012

tiny-tim

yes

and now use that value of t in a similar equation for the y direction

18. Feb 4, 2012

jackneedshelp

Ohhh kaaaaay! I get it! Thanks!

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