Question

A 500 kg car heads through a loop with a 20 m radius. At the top of the loop, the car is moving at 14 m/s. If air resistance is negligible, what would the car's velocity be at the bottom of the loop?

Answer 1

The speed of the car at the bottom of the loop is approximately 31.3 m/s

How can the velocity at the bottom of the loop be calculated?

The given parameters are;

Mass of the car = 500 kg

Radius of the loop = 20 meters

Velocity of the car at the top of the loop = 14 m/s

The kinetic energy at the top of the loop is therefore;

K E. = 0.5×500×14² = 49,000

The potential energy gained = 500×9.81×40 = 196200

Total energy = 49,000 + 196,200 = 245,200

The velocity at the bottom is therefore;

0.5×500×v² = 245,200

v² = 245,200 ÷ (0.5×500) = 980.8

v = √(980.8) ≈ 31.3

The velocity at the bottom of the loop is therefore;

• v ≈ 31.3 m/s

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