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~christina~

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## Homework Statement

Block 1 of mass 0.200 kg is sliding to the right over a frictionless elevated surface at a speed of 8.00 m / s as shown in the figure below. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N / m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off the opposite end of the elevated surface landing a distance d from the base of that surface after falling height h = 4.90 m.

(a) Write an expression that gives the displacement of block 2 as a function of time. This expression must include the values of the amplitude of vibration and the angular frequency.

(b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time. What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?

(d) What is the value of d?

http://img134.imageshack.us/img134/5667/picture2zf6.th.jpg

## Homework Equations

hm...

[tex]T= 2\pi / \omega [/tex]

## The Attempt at a Solution

I have

m1= 0.200kg

v1= 8.00m/s

T= 0.140s

h= 4.90m

d= ?

I was thinking that it's the conservation of momentum...but not sure exactly...

**a) find a exprssion that gives the displacement of block 2 as a function of time.**

Well I can find [tex]\omega [/tex] from the equation

[tex] T= 2 \pi / \omega [/tex]

[tex] \omega= 2 \pi / T [/tex]

[tex] \omega= 2 \pi / 0.140s [/tex]

[tex] \omega = 44.880rad/s [/tex]

then I can find the mass of the block m2 I think from this equation:

[tex] \omega = \sqrt{ k/ m_2} [/tex]

[tex] m_2= k/ \omega^2 [/tex]

[tex] m_2= 1208.5N/m / (44.880 rad/s)^2 [/tex]

[tex] m_2= 0.5999kg [/tex]

for the displacement vs time equation would it be this=>

[tex]x(t)= A cos(\omega t + \phi )[/tex]

A= ?

[tex] \omega= 44.880rad/s [/tex]

[tex] \phi = 0? [/tex] => I think phi is = 0 since when the block collides with the other block it's at equillibrium...thus :

x= 0

v= vi of block 1

[tex]\phi= \pm \pi/2[/tex] (here I THINK it's [tex]+ \pi /2[/tex] since it's moving to the right initially)

A= ? not sure how to find it...do I just plug in 0 for the time distance equation and find it

that way?

[tex] x(t)= A cos( 44.880rad/s t + \pi /2) [/tex] => is this correct for the displacement equation?

But how do I find A?? (I'm stumped as to how to find that)

**b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time. What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?**

well still don't know A but...

[tex]v(t)= dx/dt= A d/dt cos (44.880rad/s t + \pi/2) = -\omega A sin (44.880rad/s*t + \pi/2) [/tex]

and

[tex]a(t)= dv/dt = -\omega A d/dt sin(44.880rad/s t + \pi/2) = -\omega^2 cos(\omega*t + \pi/2)[/tex]

at t= 0.520s

I need A and I'm not sure how to find that...so I need help on that.

**d) value of d**

I think I'd find v that the block has after colliding and than use that and put that into the UAM equations to find d since I have y= 4.90 or height.

would I use the period given then plug that into the equation for velocity as a function of time then go and use that velocity as the velocity that block 1 leaves with and thus the velocity that it has as it goes off the edge?

( I would use the distance to the edge but it's not given and the surface is frictionless so that would be the same velocity as it has after being propelled by the spring back to the edge) right?

THANKS ALOT

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