A single frictionless roller coaster car of mass m=750 kg tops the first hill with speed v=15 m/s ar height h =40m as shown 1) Find the speed of the car at B and C

2) if mass m were doubled, would the speed at B increase, decrease, or remain the same?

1. The speed of the roller coaster car at point B and C: are 19.8 m/s and 0 m/s respectively.

2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on  the mass of an object.

Given the following data:

1. To find the speed of the roller coaster car at point B and C:

First of all, we would determine the potential energy of the roller coaster car by using the formula:

Where:

From the diagram, height at point B = meters

P.E = 147000 Joules.

Next, we would determine the speed by applying the law of conservation of energy.

Substituting the values, we have;

Speed, v = 19.8 m/s

From the diagram, at point C, the speed is equal to zero (0) meters per seconds.

2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on  the mass of an object.

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1.) 19.8 m/s

2.) The speed will remain the same

Explanation:

Given that a single frictionless roller coaster car of mass m = 750 kg tops the first hill with speed v = 15 m/s at height h = 40m.

To Find the speed of the car at B and C, the height = 40/2 = 20m

Let us first calculate the potential energy P. E = mgh

P.E = 750 × 9.8 × 20

P.E = 147000 J

According to conservative of energy

K.E = P.E at the bottom of the hill

1/2mV^2 = 147000

750 × V^2 = 294000

V^2 = 294000/750

V^2 = 392

V = sqrt ( 392 )

V = 19.8 m/s

Therefore, the speed of the car at point B and C will be 19.8 m/s approximately.

2) if mass m were doubled, would the speed at B increase, decrease, or remain the same?

From the conservative of energy, the energy will not depend on mass. Therefore, the speed will remain the same.

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