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=
(−1mC1⋅r)⋅vx
(b) What is the 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)=
u⋅e−tmC1⋅r
(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=
(f) What value does the horizontal component of the ball's velocity approach as t becomes infinitely large? express your answer in terms of C1, r, g, m and u as needed:
(g) What value does the vertical component of the ball's velocity approach as t becomes infinitely large? express your answer in terms of C1, r, g, m and u as needed
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)
(a) The component of the acceleration in the x direction, ax, can be given as:
ax = (-1 * m * C1 * r) * vx
(b) The acceleration in the y direction, ay, can be given as:
ay = -g + (-1 * m * C1 * r) * vy
(c) To find the expression for the horizontal component of the ball's velocity, vx(t), we can integrate the acceleration in the x direction with respect to time:
vx(t) = u * e^(-t * m * C1 * r)
(d) To find the expression for the vertical component of the ball's velocity, vy(t), we can integrate the acceleration in the y direction with respect to time:
vy(t) = -g * t + (u - m * C1 * r) * (1 - e^(-t * m * C1 * r))
(e) To determine the time it takes for the vertical speed to reach 99% of its maximum value, we can equate vy(t) to 0.99 times the maximum vertical velocity and solve for t:
0.99 * (u - m * C1 * r) = -g * t + (u - m * C1 * r) * (1 - e^(-t * m * C1 * r))
Solving this equation for t will give us the time.
(f) As t becomes infinitely large, the horizontal component of the ball's velocity, vx(t), approaches the initial horizontal velocity, u.
(g) As t becomes infinitely large, the vertical component of the ball's velocity, vy(t), approaches the terminal velocity, given by:
v_terminal = (u - m * C1 * r) / (m * C1 * r)
If you need the numerical values, you will have to substitute the given values for C1, r, g, m, and u into the expressions.
To solve the given questions, let's go step by step:
(a) The component of acceleration in the x-direction can be found by using Newton's second law, F = ma. Here, the force acting in the x-direction is the drag force, which is given by F = -C1rvx. Thus, we have:
F = ma_x
-C1rvx = ma_x
Therefore, the component of acceleration in the x-direction is:
ax = (-C1rvx) / m
Answer: ax = (-C1rvx) / m
(b) Similar to part (a), the component of acceleration in the y-direction can be found by using Newton's second law. Here, the only force acting in the y-direction is the gravitational force, which is given by F = mg. Therefore, we have:
mg = ma_y
Solving for acceleration gives:
ay = g
Answer: ay = g
(c) To find the horizontal component of the ball's velocity as a function of time, we can integrate the acceleration function from part (a) with respect to time.
ax = (-C1rvx) / m
Integrating with respect to time gives:
∫ax dt = ∫(-C1rvx) / m dt
vx(t) = vx(0) - (C1rvx/m)t
Since the initial vertical velocity is u and vy(0) = 0, we have:
vx(t) = u - (C1ru/m)t
Answer: vx(t) = u - (C1ru/m)t
(d) Similarly, to find the vertical component of the ball's velocity as a function of time, we can integrate the acceleration function from part (b) with respect to time.
ay = g
Integrating with respect to time gives:
∫ay dt = ∫g dt
vy(t) = vy(0) + gt
Since vy(0) = 0, we have:
vy(t) = gt
Answer: vy(t) = gt
(e) To find the time it takes for the vertical speed to reach 99% of its maximum value, we can set vy(t) = 0.99vy(max) and solve for t.
0.99vy(max) = gt
t = (0.99vy(max)) / g
Answer: t = (0.99vy(max)) / g
(f) As t becomes infinitely large, the horizontal component of the ball's velocity approaches the initial velocity in the x-direction, which is u.
Answer: The horizontal component of the ball's velocity approaches u.
(g) As t becomes infinitely large, the vertical component of the ball's velocity approaches the terminal velocity. In this case, since the vertical acceleration is g, the terminal velocity is given by:
Terminal velocity (v_terminal) = sqrt(mg / (C1r))
Answer: The vertical component of the ball's velocity approaches the terminal velocity, v_terminal = sqrt(mg / (C1r)).