Oscillation (conservation)

1. Dec 6, 2006

PascalPanther

1. The problem statement, all variables and given/known data
2. Relevant equations
E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
omega = Sqrt(k/m)
f = omega/(2pi) = 1/(2pi)*Sqrt(k/m)

3. The attempt at a solution
m = 150kg
M = 150kg + 50kg = 200kg
x = 15ft = 4.57m
h1 = 3.04 m
h_real = ?
x = A?

mgh = (1/2)Mv^2 + (1/2)kx^2
Since there is no velocity given, I use:
mgh = (1/2)kA^2

I need to find k:
3 cycles every 10 seconds
f = 0.3 cycles/s

f = (1/2pi)Sqrt(k/M)
0.3 = (1/2pi)Sqrt(k/M)
k = 710 N/m

(150kg)(9.8m/s^2)(h) = (1/2)(710N/m)(4.57m)^2
h = 5.04m

h_real > h1

Did I do that correctly?

2. Dec 6, 2006

Aero

Almost: you have a rounding error in the final answer.

3. Dec 6, 2006

Staff: Mentor

You have assumed that mechanical energy is conserved during the collision. Rethink that assumption, given that the car and block stick together.

Good.

4. Dec 6, 2006

PascalPanther

Hmm...
Before the collision, the block would be:
mgh = (1/2)mv^2
Then
mv = Mv_2
Is there a way to use this relationship without velocity data (actual height)?

5. Dec 6, 2006

Staff: Mentor

Sure. Start with the SHM. That should enable you to find the post-collision speed, then use that relationship (conservation of momentum) to find the pre-collision speed. Then you can deduce the height.

6. Dec 6, 2006

PascalPanther

(1/2)Mv^2 = (1/2)kA^2 right?

v = Sqrt(k/M)A
v = Sqrt(710/200kg) * (4.57m)
v = 8.6 m/s

mv = Mv
(150kg)v = (200kg)(8.6 m/s)
v= 11.5 m/s

mgh = (1/2)mv^2
(150kg)(9.8)h = (1/2)(150kg)(11.5m/s)^2
h = 6.75 m?

7. Dec 6, 2006

Looks good.