# Physics..=> drwls - I need your help

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 =)

1. 👍
2. 👎
3. 👁
1. 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

1. 👍
2. 👎
2. 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.

1. 👍
2. 👎
3. 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

1. 👍
2. 👎
4. 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)?

1. 👍
2. 👎

## Similar Questions

1. ### Physics

Block 1, of mass m1 = 0.650kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ

2. ### physics

A 1-kg block is lifted vertically 1m by a boy. The work done by the boy would be one jule or would no work be done. I'm confused, I thought lifting wasn't considered work. The work done is actually about 9.81 joules. Think of

3. ### science

which of the following distinguishes living things from nonliving things? A. living things are made up of cells; nonliving things are not** B. living things are made up of atoms; nonliving things are not C. living things are made

4. ### Physics

Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following. a. The acceleration of the 4

1. ### Physics Dynamics

A 4-kg block is connected by means of a massless rope to a 2-kg block. What is the magnitude of the acceleration if the coefficient of kinetic friction between the 4-kg block and the surface is 0.20? Note : The 4kg block is on the

2. ### Physics

A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.300, and the surface on which the 8.00- kg block rests is frictionless. A constant

3. ### Physics

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The surfaces of the blocks are rough. A constant force of magnitude F is applied to the first

4. ### Physics

If the coefficients of static and kinetic friction between a 20 kg block that sits on top of a 100 kg block are both essentially the same value of 0.5, determine the acceleration of each block for a force pulling on A of 60 N.

1. ### Physics

A block is sliding down a friction compensated runway .Which of the following statements is/are correct? 1)the speed of the block is increasing 2)the normal reaction acting by the runway on the block is increasing 3)the net force

2. ### Physics Check My Work Please

A 8 kg block is moving with an initial speed of 6 m/s on a rough horizontal surface. If the force is 12 N, approximately how far does the block travel before it stops? *I have: V^2=Vi^2+2ax 0=(6 m/s)^2 + 2(-9.8 m/s^2)(x) -36

3. ### Pre-algebra

Dwayne walks a total of 960 feet around 3 sides of a square city block. Which equation and solution show a correct way to find the perimeter, P, of the block?

4. ### Physics

Block A sits on a horizontal tabletop. There is friction between the surface and Block A. The string passes over a frictionless, massless pulley. Block B hangs down vertically. When the two blocks are released, Block B accelerates