A roller-coaster car speeds down a hill past point A where R1 = 8.6 m and then rolls up a hill past point B where R2 = 16.0 m.

(a) The car has a speed of 18.4 m/s at point A. if the track exerts a normal force on the car of 2.21 104 N at this point, what is the mass of the car?

(b) What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?

a) The mass of the car can be calculated using the equation F = ma, where F is the normal force, m is the mass of the car, and a is the acceleration. Rearranging the equation to solve for m, we get m = F/a. Substituting the given values, we get m = 2.21 x 104 N / (18.4 m/s2) = 1.2 x 103 kg.

b) The maximum speed the car can have at point B for the gravitational force to hold it on the track can be calculated using the equation F = mg, where F is the gravitational force, m is the mass of the car, and g is the acceleration due to gravity. Rearranging the equation to solve for v, we get v = √(2Rg), where R is the radius of the hill. Substituting the given values, we get v = √(2 x 16.0 m x 9.8 m/s2) = 20.2 m/s.

(a) To solve for the mass of the car at point A, we can use the equation:

N - mg = mv² / R

where N is the normal force, m is the mass of the car, g is the acceleration due to gravity, v is the speed of the car, and R is the radius of the curve at point A.

Given:
N = 2.21 x 10^4 N
v = 18.4 m/s
R = 8.6 m
g = 9.8 m/s²

Substituting these values into the equation, we have:

2.21 x 10^4 - m(9.8) = m(18.4² / 8.6)

Simplifying and solving for m, we get:

2.21 x 10^4 - 9.8m = 37.8m

Combining like terms, we have:

47.6m = 2.21 x 10^4

Dividing both sides by 47.6, we find:

m ≈ 464.28 kg

Therefore, the mass of the car is approximately 464.28 kg at point A.

(b) To determine the maximum speed the car can have at point B to stay on the track, we need to consider the minimum centripetal force required to overcome the gravitational force.

At point B, the only vertical force acting on the car is the gravitational force, which is given by:

F_gravity = mg

To keep the car on the track, the centripetal force should also be pointing towards the center of the curve and at least equal to the gravitational force. So, we have:

F_centripetal = mg

According to the formula for centripetal force:

F_centripetal = mv² / R

Setting the centripetal force equal to the gravitational force, we can solve for v:

mv² / R = mg

Dividing both sides by m and multiplying by R, we get:

v² = gR

Taking the square root of both sides:

v = √(gR)

Substituting the given values:

g ≈ 9.8 m/s²
R = 16.0 m

v = √(9.8 * 16.0) ≈ √(156.8) ≈ 12.53 m/s

Therefore, the maximum speed the car can have at point B to stay on the track is approximately 12.53 m/s.

To solve these questions, we can use the principles of circular motion and the equation for the centripetal force acting on an object moving in a circle.

(a) To find the mass of the car at point A, we can use the equation for centripetal force:

Fc = (m * v^2) / R

Where Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and R is the radius of the circular path (in this case, R1).

Given that Fc = 2.21 * 10^4 N, v = 18.4 m/s, and R1 = 8.6 m, we can rearrange the equation to solve for m:

m = (Fc * R) / v^2

Plugging in the given values:

m = (2.21 * 10^4 N * 8.6 m) / (18.4 m/s)^2

Calculating this:

m ≈ 764.89 kg

Therefore, the mass of the car at point A is approximately 764.89 kg.

(b) To find the maximum speed the car can have at point B for the gravitational force to hold it on the track, we need to consider the minimum centripetal force required to keep the car on the track at point B. At this point, the car is at the top of the hill, so the gravitational force is acting downwards and has an opposite direction to the centripetal force.

The centripetal force required is provided by the gravitational force:

Fc = Fg

Using the equation for gravitational force:

Fc = (m * v^2) / R

Where Fc is the centripetal force, Fg is the gravitational force, m is the mass of the car, v is the speed of the car, and R is the radius of the circular path (in this case, R2).

Rearranging the equation to solve for v:

v = √(Fg * R / m)

Given that R2 = 16.0 m and we already found the mass of the car in part (a) to be approximately 764.89 kg, we can calculate the maximum speed:

v = √(9.8 m/s^2 * 16.0 m / 764.89 kg)

Calculating this:

v ≈ 6.01 m/s

Therefore, the maximum speed the car can have at point B for the gravitational force to hold it on the track is approximately 6.01 m/s.