Question

At low speeds (especially in liquids rather than gases), the drag force is proportional to the speed rather than it's square, i.e., F = -c_1*r*v , where c_1 is a constant. At time t = 0, a small ball of mass is projected into a liquid so that it initially has a horizontal velocity of in the direction as shown. The initial speed in the vertical direction ( y) is zero. The gravitational acceleration is g. Consider the cartesian coordinate system shown in the figure (+x to the right and +y downwards).

Express the answer of the following questions in terms of some or all of the variables c_1, r , m , g, v_x , v_y , and u (enter C_1 for , v_x for and v_y for ). Enter e^(-z) for exp(-z) (the exponential function of argument -z).

(a) What is component of the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_x , c_1 , r , g , m and u as needed: What is the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_y , c_1 , r , g, m and as needed: Using your result from part (a), find an expression for the horizontal component of the ball's velocity as a function of time? Using your result from part (b), find an expression for the vertical component of the ball's velocity as a function of time t. How long does it take for the vertical speed to reach 99 of its maximum value.What value does the horizontal component of the ball's velocity approach as t becomes infinitely large?What value does the vertical component of the ball's velocity approach as t becomes infinitely large?

Answer 1

a)(-C_1*r*v_x)/m
b)g-(C_1*r*v_y)/m
c)u*e^(-(C_1*r*t)/m)
d)((m*g)/(C_1*r))-((m*g)/(C_1*r))*e^(-(C_1*r*t)/m)
e)4.6*(m/(C_1*r))
f)0
g)(m*g)/(C_1*r)

Answer 2

U r awesome...

Answer 3

You are welcome, did you get number 6?

Answer 4

q6 (5*R)/2

did anyone got vertical spring?

Answer 5

Darn, I said 6, I meant 5 the half loop. Vertical spring is a) .4785
b) .72
c) 1.1785
d) 1.812

Answer 6

for me q6 is full loop and q5 is half loop
i did not got that one i am on half way
your answers are showing wrong for me.
do you have the steps?

Answer 7

You mean the one that starts d1=.4785? I took the general equation, Acos(wt+p), used the initial conditions to solve for the constants. Then I used that to get the answers. At least that is what I remember. I don't recall all the details I'll see if I can find the details.

Answer 8

yup

Answer 9

I'm looking for my notes. The first one comes from the equation ma=-kx. then
(ma/k)=-x = (4*9.81)/82=x=.4785

Answer 10

The second one comes from
f=(1/2pi)sqrt(k/m)
=(1/2pi)*sqrt(82/4)=.72

Answer 11

for the other two I think I got
x(t)=.4cos(4.53t+pi) and then set v(t)=0 to get the time and then used that to get the velocity. And I used v(tm)=0 to get the time when it reaches the highest point and plugged that value back into x(t).

Answer 12

thnx..
and what about spring block question?

Answer 13

I got 3, don't ask me how I got it. I'm not sure.

Answer 14

just tell the answer

Answer 15

The answer is "3"

Answer 16

Hey are you in quantum also?

Answer 17

yes