For this problem, you are allowed to consult with other members of the class, but not with faculty and not with anyone outside the course. Simplify all your expressions.
A car of mass, 2m, is going around a turn at a constant radius, R, with velocity, v0. The turn is not banked – it is a flat surface. Assume the car and the road are on Earth’s surface.
a) Is the friction that holds the car on the road around the turn static or kinetic? Explain your answer.
b) Find the minimum coefficient of friction needed for the car to make the turn in terms of v0, R, g, and any other variables or constants that are necessary for the expression.
c) Find the kinetic energy of the car in terms of m, v0, and any constants.
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d) The car hits an icy patch and careens off the turn in a straight line. It maintains its velocity, v0, until it hits a block of mass, m, attached to a massless spring with spring constant, k. The car has a perfectly inelastic collision with the block attached to the spring. Find the velocity, v2, of the combined car and block immediately after the collision takes place in terms of m, v0, and any needed constants.
e) Determine the distance that the spring compresses after the collision in terms of m, k, v0, and any necessary constants.
f) Find the time for the spring to return to its original position the first time after it compresses and starts to stretch in terms of m, k, and any other necessary variables or constants.
What happens to your answer in part f if the car has a mass of 11m instead?
a) To determine whether the friction holding the car on the road around the turn is static or kinetic, we need to consider the conditions required for each type of friction.
Static friction occurs when two surfaces are not slipping against each other. In this case, the car is going around a turn at a constant velocity, which suggests that there is no slipping between the tires and the road. Therefore, the friction holding the car on the road is static.
b) To find the minimum coefficient of friction needed for the car to make the turn, we can analyze the forces acting on the car. In this case, the centripetal force required to keep the car moving in a circle is provided by the frictional force between the tires and the road.
The centripetal force can be calculated using the equation: Fc = mv0^2/R, where m is the mass of the car, v0 is the velocity of the car, and R is the radius of the turn.
Since the frictional force provides the centripetal force, we can equate the two: Ff = Fc.
The frictional force can be written as Ff = μmg, where μ is the coefficient of friction and g is the acceleration due to gravity.
Setting Ff equal to Fc, we can solve for the minimum coefficient of friction (μ) required: μmg = mv0^2/R. Simplifying the equation, we get: μ = v0^2/(gR).
Therefore, the minimum coefficient of friction needed for the car to make the turn is given by μ = v0^2/(gR).
c) The kinetic energy of the car can be calculated using the formula: KE = (1/2)mv0^2, where m is the mass of the car and v0 is the velocity of the car.
Therefore, the kinetic energy of the car is KE = (1/2)mv0^2.
d) After the collision with the block attached to the spring, the car and the block become a combined system moving with a new velocity, v2. To find v2, we can apply the principle of conservation of momentum.
Before the collision, the momentum of the car is given by p1 = mv0, where m is the mass of the car and v0 is the velocity of the car.
After the collision, the momentum of the combined car and block system is given by p2 = (m + m)v2, where m is the mass of the car, m is the mass of the block, and v2 is the velocity of the combined system.
Since momentum is conserved, we can equate p1 and p2: mv0 = 2mv2.
Simplifying the equation, we can solve for v2: v2 = v0/2.
Therefore, the velocity of the combined car and block immediately after the collision is v2 = v0/2.
e) To determine the distance that the spring compresses after the collision, we can use the concept of conservation of mechanical energy.
The initial mechanical energy of the system is given by the kinetic energy of the car before the collision, which is KE = (1/2)mv0^2.
The final mechanical energy of the system is given by the potential energy stored in the compressed spring. This can be expressed as PE = (1/2)kx^2, where k is the spring constant and x is the compression distance of the spring.
Setting the initial mechanical energy equal to the final mechanical energy, we can solve for x: (1/2)mv0^2 = (1/2)kx^2.
Simplifying the equation, we get: x = sqrt((mv0^2) / k).
Therefore, the distance that the spring compresses after the collision is x = sqrt((mv0^2) / k).
f) To find the time for the spring to return to its original position the first time after it compresses and starts to stretch, we can use the formula for the period of an oscillating spring.
The period of an oscillating spring is given by T = 2π(sqrt(m/k)), where m is the mass attached to the spring and k is the spring constant.
Therefore, the time for the spring to return to its original position the first time after it compresses and starts to stretch is T = 2π(sqrt(m/k)).
If the car has a mass of 11m instead, the period of the spring's oscillation will remain the same. The mass of the car does not affect the period of the oscillation if the spring constant remains constant. Therefore, the answer in part f will not change.