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:
ax= (-C_1*r*v_x)/m
To solve this problem, we will use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
(a) The drag force acting on the ball in the x-direction is given by the equation F⃗ = −C1rv⃗ . Since the drag force is in the opposite direction of the velocity, it will cause deceleration. Thus, the acceleration in the x-direction can be given as:
F_x = -C1 * r * v_x
Using Newton's second law, we equate this force to the mass of the object multiplied by its acceleration:
m * a_x = -C1 * r * v_x
Solving for a_x, we get:
a_x = -(C1 * r * v_x) / m
Therefore, the component of acceleration in the x-direction is ax = -(C1 * r * vx) / m.
(b) The acceleration in the y-direction is due to gravity and the drag force acting in the y-direction, which is given by F⃗ = −C1rv⃗ . Thus, the net force in the y-direction is:
F_y = -C1 * r * v_y + mg
Using Newton's second law:
m * a_y = -C1 * r * v_y + mg
Solving for a_y, we get:
a_y = (-C1 * r * v_y + mg) / m
Therefore, the component of acceleration in the y-direction is ay = (-C1 * r * vy + mg) / m.
(c) The horizontal component of the ball's velocity can be found by integrating the acceleration in the x-direction with respect to time. Integrating a_x = -(C1 * r * vx) / m yields:
vx(t) = -(C1 * r * u / m) * t + u
Therefore, the expression for the horizontal component of the ball's velocity as a function of time t is vx(t) = -(C1 * r * u / m) * t + u.
(d) The vertical component of the ball's velocity can be found by integrating the acceleration in the y-direction with respect to time. Integrating a_y = (-C1 * r * vy + mg) / m yields:
vy(t) = -(C1 * r * u / m) * (1 - e^(-(C1 * r * t) / m)) - (g * t)
Therefore, the expression for the vertical component of the ball's velocity as a function of time t is vy(t) = -(C1 * r * u / m) * (1 - e^(-(C1 * r * t) / m)) - (g * t).
(e) To find the time it takes for the vertical speed to reach 99% of its maximum value, we set vy(t) equal to 0.99 times its maximum value and solve for t:
0.99 * (-C1 * r * u / m) * (1 - e^(-(C1 * r * t) / m)) - (g * t) = 0
Solving this equation for t will give us the time at which the vertical speed reaches 99% of its maximum value. Note that we do not need the value of C1 to find this time.