A rollercoaster cart leaves point A at a speed of 4 m/s, passes through point B on the ground, and reaches point C at a speed of 6 m/s. The total mass of the cart assembly and the passengers are 300 kg, (a) What is the speed of the cart at point B? (b) What is the height of point A?
Answers & Comments
To solve this problem, we can apply the principle of conservation of mechanical energy. The total mechanical energy of the rollercoaster cart at any point along its path remains constant, assuming no external forces (such as friction) are acting on it.
Let's denote the speed at point A as \( v_A \), the speed at point B as \( v_B \), and the speed at point C as \( v_C \). The height of point A will be denoted as \( h_A \).
(a) To find the speed of the cart at point B, we can equate the mechanical energies at points A and B:
\(\frac{1}{2} m v_A^2 + mgh_A = \frac{1}{2} m v_B^2 + mgh_B\)
Since the rollercoaster starts from rest at point A, \( v_A = 0 \). Also, the height at point B is the same as at point A, so \( h_A = h_B \). Substituting these values, the equation becomes:
\(0 + mgh_A = \frac{1}{2} m v_B^2 + mgh_B\)
Simplifying and canceling out the mass \( m \), we get:
\(gh_A = \frac{1}{2} v_B^2 + gh_B\)
Since we want to find the speed at point B, we can rearrange the equation as:
\(v_B^2 = 2gh_A - 2gh_B\)
Substituting the given values, we have:
\(v_B^2 = 2 \cdot 9.8 \cdot h_A - 2 \cdot 9.8 \cdot h_B\)
(b) To find the height of point A, we can use the conservation of mechanical energy equation between points A and C:
\(\frac{1}{2} m v_A^2 + mgh_A = \frac{1}{2} m v_C^2 + mgh_C\)
Since \( v_A = 0 \) and \( v_C = 6 \, \mathrm{m/s} \), the equation becomes:
\(0 + mgh_A = \frac{1}{2} m (6 \, \mathrm{m/s})^2 + mgh_C\)
Simplifying and canceling out the mass \( m \), we get:
\(gh_A = 18 + gh_C\)
Since the rollercoaster starts from the ground at point B, \( h_C = 0 \). Substituting this value, the equation becomes:
\(gh_A = 18 + 0\)
Simplifying further, we have:
\(gh_A = 18\)
Now, we can solve for \( h_A \) by dividing both sides of the equation by \( g = 9.8 \, \mathrm{m/s^2} \):
\(h_A = \frac{18}{9.8} \, \mathrm{m}\)
Simplifying, we find:
\(h_A \approx 1.84 \, \mathrm{m}\)
Therefore, the speed of the cart at point B is \( \sqrt{2 \cdot 9.8 \cdot 1.84} \, \mathrm{m/s} \), and the height of point A is approximately \( 1.84 \, \mathrm{m} \).
Point A is 1.84
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