A roller coaster is designed such that riders experience ''weightlessness'' as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between

Question

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kg/s with a constant relative speed of 11 km/s. What is the acceleration of the rocket one minute after the blast?

Answer 1

To solve this problem, we need to determine at what speed a roller coaster car can maintain a state of "weightlessness" at the top of a hill with a radius of curvature of 20 m. Weightlessness in this context means that the normal force acting on the riders is zero.

In physics, weightlessness in a roller coaster at the top of a circular path implies that the only force acting on the riders is gravity. This happens when the centripetal force required for circular motion is exactly provided by the weight of the riders. Mathematically, this can be expressed as:

mg = \frac{mv^2}{r}

where:

m is the mass of the riders or the car,
g is the acceleration due to gravity (approximated as 9.81 \, \text{m/s}^2),
v is the velocity at the top,
r is the radius of curvature of the hill (20 m).

Canceling m from both sides of the equation and solving for v, we get:

g = \frac{v^2}{r}

Multiplying both sides by r gives:

v^2 = gr

Taking the square root of both sides, the expression for v becomes:

v = \sqrt{gr}

Inserting the known values for g and r:

v = \sqrt{9.81 \times 20}

Calculating further, we find:

v = \sqrt{196.2} \approx 14.004 \, \text{m/s}

Therefore, the speed of the car at the top of the hill should be approximately 14 m/s, which falls within the given range of 14 m/s and 15 m/s.

Hence, the correct answer is the option: 14 m/s and 15 m/s.

Answer 2