Physics..=> drwls  I need your help
posted by ~christina~
I need help seeing if my thoughts are correct and how to do some things.
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. 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.
I came up with
x(t)= Acos(omega*t + pi/2)
not sure about +/ for the angle though and how you know which sign to have.(need help on this determination)
I found omega and m2 through: T= 2pi/omega= 2pi sqrt(m/k)
m1v1i + m2v2i = m1v1f + m2v2f
and found v2f and v1f
I was thinking that the v1f I found was the same velocity that the block 1 leaves with and travels off the table with IS THIS CORRECT?
Then I was thinking of using the v1f and v2f in energy equation to find the distance that the spring compresses (Amplitude) so I can plug it into the equation for cos
1/2mv1f + 1/2m2vf = 1/2kx^2
Is this alright?
b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time.
once again I'm not sure if phi's angle or even if phi is correct.
x(t)= A cos (omega*t + pi/2)
v(t)= omega A sin( omega*t + pi/2)
a(t)= omega^2 cos (omega*t + pi/2)
c) What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?
I think I'd just plug into the equation after I find the values from a
(d) What is the value of d?
I know this is projectile motion problem with I think v in x direction...but if it is then would an angle be included? I think yes but I haven't worked with many problems with a object falling after sliding off a level surface.
how would I approach this?
Thank you drwls =)

drwls
a) The displacement of block 2 will be 0 at t=0, and it will vibrate about this position. They already tell you the period is P = 0.140 s. Figure out tne mass m2 from the relation
P = 2 pi sqrt (m2/k) = 0.140 s
m2/k = [P/(2 pi)]^2 = 4.965*10^4.
I get m2 = 0.600 kg, 3 times the mass of m1.
You need to compute the amplitude of vibration to complete this part. You can get that by using energy and momentum conservation to compute the velocity of m2 right after collision. I believe you will find this to be 4.00 m/s. Vmax of mass2 equals omega*A, where A is the amplitude. In your case,
omega = 2 pi/P = 44.88 rad/s
Therefore A = (4.00 m/s)/44.88 rad/s) = 8.91*10^2 m
Another way to get the amplitude is to set (1/2) k A^2 equal to the kinetic energy of m2 right after the collision, which you suggested. It should give the same answer. Try it and see.
The displacement equation for mass 2 is, if I'm right,
X = 8.91*10^2 m * sin(2 pi t/P)
= 8.91*10^2 sin (44.88 t)
2) This step is straightforward since you know know X(t)  assuming I did the calculations correctly. This you need to verify.
3) Yes, just plug in the numbers. Since that is exactly 3 periods later, you should get the same values you had at t=0
4) In doing the elastic collision problem, you should find that mass 1 bpunces back with a velocity of 4.0 m/s. The time it takes to fall a veritcal height of 4.90 m is
t = sqrt (2H/g) = sqrt 1 = 1.00 s
The distance d will therefore be 4.0 m from the base 
~christina~
X = 8.91*10^2 m * sin(2 pi t/P)
= 8.91*10^2 sin (44.88 t)
I don't get this...why did you use sin? and how did you get 2pi t/P ??
I used these formulas below but are they incorrect?
x(t)= A cos (omega*t + pi/2)
v(t)= omega A sin( omega*t + pi/2)
a(t)= omega^2 cos (omega*t + pi/2)
P.S. I was also wondering if there is a good site on the web that can explain the relation of the position of a spring to the sin/cos function since I have problems visualizing the two and which one I should use for which situation. 
drwls
I used sin because the displacement at time =0, and then it becomes positive at first.
cos (wt + pi/2) is the same thing as sin wt, anyway.
Your formula is OK if you define positive motion to be opposite to the direction of m1 before impact.
However, in the first part I think they want you to provide the actual value of A.
2 pi t/P is the same thing as w t, since 1/P is the frequency f and
2 pi f = w 
~christina~
Thanks drwls
I was wondering though if I used sin then wouldn't the
v(t)= omega A cos (omega* t)?
and
a(t)= omega^2 A sin(omega* t)?
Respond to this Question
Similar Questions

Physics
A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a light, frictionless pulley. The 30.0 kg block is connected to a spring that has negligible mass and a force constant of 200 N/m. The spring is unstretched … 
physics
Block A has a mass of 4.1 kg, and it is on a frictionless surface. A string tied to this mass passes over a frictionless pulley, and is attached to a hanging block (Block B), as shown in the image below. The blocks then accelerate … 
Physics
A 2.9 kg block slides along a frictionless surface at 1.3 m/s. A second block, sliding at a faster 5.0 m/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.7 m/s. What was the mass … 
physics
Pleaseee help explain how to draw the free body diagrams for this problem!!! Two blocks with masses m1 = 10.9 kg and m2 = 79.3 kg, shown in the figure (in the figure the 79.3kg block(m2) is on a flat surface, the 10.9kg(m1) block is … 
Physics Help Please
A horizontal spring with a spring constant of 200.0 N/m is compressed 25.0 cm and used to launch a 3.00 kg block across a frictionless surface. After the block travels some distance, the block goes up a 32 degree incline that has a … 
Physics
I have posted the same questions over and over and received no help. Will someone please help me? 
Physics
The figure below shows block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surface at a speed of 5.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring … 
Physics
A bullet with a velocity of 200 m/s is shot at a 2kg block resting on a horizontal frictionless surface. The bullet imbeds in the block and the bulletblock system slides across the surface with a speed of 5 m/s. A) what is the mass … 
physics
Two blocks are connected by a lightweight cord that passes over a frictionless, massless pulley. The bottom block has a mass of 2.00 kg and the top block has mass 0.500 kg. The coefficient of kinetic friction at each sliding surface … 
physics
home / study / science / physics / questions and answers / a block of mass 3kg s sliding along the frictionless ... Question A block of mass 3kg s sliding along the frictionless horizontal surface with a speed of 2m/s. 1. What is the …