Challenge Problem: A wrecking ball is suspended from a 14 m long cable that makes a 360 angle with the vertical. The ball is released and swings down. What is the ball’s speed at the lowest point (g = 9.81ms-2) ? Hint: GPEmax = KEmax, you will also need to use trigonometry to help you solve this problem. Answer to 1 decimal place and only put in the numerical value. Do not write the unit in your answer.

This is a short-answer question. Show all working on paper, including the relevant equations and units.

Question

Challenge Problem: A wrecking ball is suspended from a 14 m long cable that makes a 360 angle with the vertical. The ball is released and swings down. What is the ball’s speed at the lowest point (g = 9.81ms-2) ? Hint: GPEmax = KEmax, you will also need to use trigonometry to help you solve this problem. Answer to 1 decimal place and only put in the numerical value. Do not write the unit in your answer.

This is a short-answer question. Show all working on paper, including the relevant equations and units.

BlackTom AI · Accepted Answer

First, I restate the problem to ensure understanding: a wrecking ball is suspended from a 14 m long cable, the cable makes a 36° angle with the vertical, the ball is released and swings to the lowest point, and we’re to find its speed at the bottom using g = 9.81 m/s^2. The hint tells us GPEmax = KEmax, and that trigonometry will be needed.

Step 1: Identify the energy conversion. At the highest point (initial release) the ball has gravitational potential energy relative to the lowest point, and no kinetic energy (assuming it is released from rest). At the lowest point, all of that gravitational potential energy has been converted to kinetic energy. Therefore, m g h = 1/2 m v^2, where h is the vertical drop from the initial position to the lowest point. The mass m cancels, simplifying the relation to v = sqrt(2 g h).

Step 2: Compute h using geometry. The vertical drop h is the difference in vertical height between the initial position (at angle 36° from vertical) and the bottom (the lowest point). For a pendulum of length L, the vertical height above the bottom is h = L - L cos(theta) = L (1 - cos(theta)). Here L = 14 m and theta = 36°. Evaluate cos(36°) ≈ 0.8090. Therefore h ≈ 14 * (1 - 0.8090) ≈ 14 * 0.1910 ≈ 2.674 m.

Step 3: Substitute into v = sqrt(2 g h). With g = 9.81 m/s^2 and h ≈ 2.674 m, compute the product inside the square root: 2 * g * h ≈ 2 * 9.81 * 2.674 ≈ 52.4. Then v ≈ sqrt(52.4) ≈ 7.24 m/s.

Step 4: Round to one decimal place as required. 7.24 m/s rounds to 7.2 m/s when expressed to one decimal place. Since the problem asks for the numerical value without units, the expected numeric answer would be 7.2.

Notes on potential discrepancies: The provided sample answer 8.8 does not match the calculation using the given values (L = 14 m, theta = 36°, g = 9.81 m/s^2). If any of these inputs were different (for example, a different angle, a different height, or a different length), the result would change accordingly. Also, ensure the angle is indeed 36° with the vertical; if it were 36° with the horizontal, cos would be sin(36°) in the height formula, leading to a different h and a different v.

In summary, following the standard energy conservation approach with the height h = L(1 - cos 36°) yields v ≈ 7.2 (to one decimal place).

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