Mine Shaft Kinematics: Understanding Particle and Cage Motion

This problem involves analyzing the motion of two separate objects: a descending cage in a mine-shaft and a particle dropped from the top of the shaft. The key is to determine the time it takes for the particle to hit the cage, considering their different initial conditions and acceleration values.

Cage Motion Analysis Before Particle Drop

First, let's understand the state of the cage at the moment the particle is dropped. The cage starts from rest and descends with a constant acceleration of \(2 \text{ ft/s}^2\). The particle is dropped after the cage has been in motion for \(10\) seconds.

• Initial velocity of the cage (\(u_c\)): \(0 \text{ ft/s}\)
• Acceleration of the cage (\(a_c\)): \(2 \text{ ft/s}^2\)
• Time elapsed before the particle is dropped (\(t_0\)): \(10 \text{ seconds}\)

We can calculate the distance covered by the cage and its velocity at \(t_0 = 10 \text{ seconds}\) using the equations of motion:

1. Displacement of the cage in the first 10 seconds (\(s_{c0}\)):

Using the formula \(s = ut + \frac{1}{2}at^2\):

\(s_{c0} = u_c t_0 + \frac{1}{2}a_c t_0^2\)

\(s_{c0} = (0 \text{ ft/s})(10 \text{ s}) + \frac{1}{2}(2 \text{ ft/s}^2)(10 \text{ s})^2\)

\(s_{c0} = 0 + \frac{1}{2}(2)(100)\)

\(s_{c0} = 100 \text{ ft}\)

2. Velocity of the cage at 10 seconds (\(v_{c0}\)):

Using the formula \(v = u + at\):

\(v_{c0} = u_c + a_c t_0\)

\(v_{c0} = 0 \text{ ft/s} + (2 \text{ ft/s}^2)(10 \text{ s})\)

\(v_{c0} = 20 \text{ ft/s}\)

So, when the particle is dropped, the cage is \(100 \text{ ft}\) below the top of the shaft and moving downwards at \(20 \text{ ft/s}\).

Particle and Cage Motion Relative to Dropping Time

Let's consider the time \(t\) as the duration *after the particle is dropped* until it hits the cage. We will set our reference point (origin) at the top of the shaft, with the downward direction as positive.

Particle Motion Equation

The particle is dropped from the top of the shaft, meaning its initial velocity is zero. It accelerates due to gravity.

• Initial velocity of the particle (\(u_p\)): \(0 \text{ ft/s}\)
• Acceleration of the particle (\(a_p\)): \(32 \text{ ft/s}^2\) (given)

The displacement of the particle from the top of the shaft after time \(t\) is:

\(s_p(t) = u_p t + \frac{1}{2}a_p t^2\)

\(s_p(t) = (0)t + \frac{1}{2}(32)t^2\)

\(s_p(t) = 16t^2\)

Cage Motion Equation

At the moment the particle is dropped (which is \(t=0\) for our new time reference), the cage is already at \(100 \text{ ft}\) from the top and has an initial downward velocity of \(20 \text{ ft/s}\). It continues to accelerate at \(2 \text{ ft/s}^2\).

• Initial position of the cage (at \(t=0\)): \(s_{c0} = 100 \text{ ft}\)
• Initial velocity of the cage (at \(t=0\)): \(v_{c0} = 20 \text{ ft/s}\)
• Acceleration of the cage (\(a_c\)): \(2 \text{ ft/s}^2\)

The displacement of the cage from the top of the shaft after time \(t\) is:

\(s_c(t) = s_{c0} + v_{c0}t + \frac{1}{2}a_c t^2\)

\(s_c(t) = 100 + 20t + \frac{1}{2}(2)t^2\)

\(s_c(t) = 100 + 20t + t^2\)

Collision Time Calculation

The particle hits the cage when their displacements from the top of the shaft are equal. So, we set \(s_p(t) = s_c(t)\):

\(16t^2 = 100 + 20t + t^2\)

Rearrange the equation to form a standard quadratic equation (\(At^2 + Bt + C = 0\)):

\(16t^2 - t^2 - 20t - 100 = 0\)

\(15t^2 - 20t - 100 = 0\)

To simplify, divide the entire equation by \(5\):

\(3t^2 - 4t - 20 = 0\)

Now, we use the quadratic formula to solve for \(t\):

\(t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)

Here, \(A=3\), \(B=-4\), \(C=-20\).

\(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-20)}}{2(3)}\)

\(t = \frac{4 \pm \sqrt{16 + 240}}{6}\)

\(t = \frac{4 \pm \sqrt{256}}{6}\)

\(t = \frac{4 \pm 16}{6}\)

This gives two possible values for \(t\):

• \(t_1 = \frac{4 + 16}{6} = \frac{20}{6} = \frac{10}{3} \text{ seconds}\)
• \(t_2 = \frac{4 - 16}{6} = \frac{-12}{6} = -2 \text{ seconds}\)

Since time cannot be negative, we discard \(t_2 = -2 \text{ seconds}\).

Therefore, the time elapsed before the particle hits the cage is \(t = \frac{10}{3} \text{ seconds}\).

This can also be expressed as a mixed number:

\(t = 3 \frac{1}{3} \text{ seconds}\)

Time Calculation Summary Table

The problem is a classic example of relative motion and involves applying equations of motion for objects under constant acceleration. Carefully setting up the initial conditions and displacement equations for both the cage and the particle is crucial for finding the correct time of collision.