At low speeds (especially in liquids rather than gases), the drag force is proportional to the speed rather than it's square, i.e., F⃗ = −C1rv⃗ , where C1 is a constant. At time t = 0, a small ball of mass m is projected into a liquid so that it initially has a horizontal velocity of u in the +x 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 C1, r, m, g, vx, vy, u and t (enter C_1 for C1, v_x for vx and v_y for vy). Enter e^(-z) for exp(-z) (the exponential function of argument -z).
(a) What is component of the acceleration in the x direction as a function of the component of the velocity in the x direction vx? express your answer in terms of vx, C1, r, g, m and u as needed:
ax=
acceleration in the y direction as a function of the component of the velocity in the y direction vy? express your answer in terms of vy, C1, r, g, m and u as needed:
ay=
(c) Using your result from part (a), find an expression for the horizontal component of the ball's velocity as a function of time t? express your answer in terms of C1, r, g, m, u and t as needed: (enter e^(-z) for exp(-z)).
vx(t)=
(d) Using your result from part (b), find an expression for the vertical component of the ball's velocity as a function of time t? express your answer in terms of C1, r, g, m, u and t as needed: (enter e^(-z) for exp(-z)).
vy(t)=
(e) How long does it take for the vertical speed to reach 99% of its maximum value? express your answer in terms of C−1, r, g, m and u as needed:
t=
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)
To solve these problems, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
(a) The component of acceleration in the x-direction can be found by considering the forces acting on the ball in the x-direction:
ma_x = -C_1rv_x
Solving for a_x, we have:
a_x = -C_1rv_x/m
(b) The component of acceleration in the y-direction can be found by considering the forces acting on the ball in the y-direction:
ma_y = -mg - C_1rv_y
Solving for a_y, we have:
a_y = (-mg - C_1rv_y)/m
(c) To find the horizontal component of the ball's velocity as a function of time, we can integrate the acceleration with respect to time:
v_x(t) = ∫a_x dt = ∫(-C_1rv_x/m)dt = (-C_1r/m)∫v_x dt
Assuming the initial velocity v_x(0) = u, we have:
v_x(t) = (-C_1r/m)∫u dt = (-C_1r/m)ut
(d) To find the vertical component of the ball's velocity as a function of time, we can also integrate the acceleration with respect to time:
v_y(t) = ∫a_y dt = ∫((-mg - C_1rv_y)/m)dt = (-mg - C_1r/m)∫v_y dt
Assuming the initial velocity v_y(0) = 0, we have:
v_y(t) = ∫(-mg - C_1rv_y)/m dt = -(mg + C_1r∫v_y dt)/m
(e) To find the time it takes for the vertical speed to reach 99% of its maximum value, we can set v_y(t) equal to 0.99 times the maximum vertical velocity V_y (which can be found when dy/dt = 0):
0.99V_y = -(mg + C_1r∫v_y dt)/m
Solving for t, we have:
t = ∫(0.99V_y)m/(mg + C_1r∫v_y dt) dt
Note that in equations (c), (d), and (e), we have assumed that the initial velocities in the x and y-directions are u and 0, respectively. Make sure to double-check if there are any given initial conditions that might affect the equations.