A 2100 kg car traveling at 12.0 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick together and move 3.90 m before friction causes them to stop. Determine the coefficient of kinetic friction betwen the cars and the road, assuming that the negative acceleration is constant and that all wheels on both cars lock at the time of impact.
M1*V1 + M2*V2 = M1*V + M2*V
2100*12 + 2780*0 = 2100V + 2780V
25,200 = 4880V
V = 5.16 m/s.
See Related Questions: 10-30-14, 9:40 PM.
To determine the coefficient of kinetic friction between the cars and the road, we need to analyze the forces acting on the cars.
Let's start by calculating the initial momentum of the two cars. The momentum (p) is given by the product of mass (m) and velocity (v).
For the 2100 kg car:
Initial momentum (p1) = mass (m1) × velocity (v1)
= 2100 kg × 12.0 m/s
= 25200 kg·m/s
For the 2780 kg car (initially at rest):
Initial momentum (p2) = mass (m2) × velocity (v2)
= 2780 kg × 0 m/s
= 0 kg·m/s
After the collision, the two cars stick together and move with a common velocity (vf) before coming to a stop.
By applying the principle of conservation of momentum, we can find the common velocity after the collision.
The total initial momentum must equal the total final momentum.
Total initial momentum = Total final momentum
(p1 + p2) = (m1 + m2) × vf
Substituting the values we have:
(25200 kg·m/s + 0 kg·m/s) = (2100 kg + 2780 kg) × vf
25200 kg·m/s = 4880 kg × vf
Now we can solve for the final velocity (vf):
vf = 25200 kg·m/s ÷ 4880 kg
≈ 5.164 m/s
Next, we need to determine the net force acting on the cars to bring them to a stop.
The net force (F) can be calculated using Newton's second law of motion, which states that the force (F) is equal to mass (m) multiplied by acceleration (a):
F = m × a
Since the cars are brought to a stop, the acceleration (a) can be calculated using the following kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vi = initial velocity (12.0 m/s)
vf = final velocity (0 m/s)
d = displacement (3.90 m)
Rearranging the equation to solve for acceleration (a):
a = (vf^2 - vi^2) ÷ (2d)
= (0 m/s - 12.0 m/s)^2 ÷ (2 × 3.90 m)
= (-12.0 m/s)^2 ÷ (2 × 3.90 m)
= 144 m^2/s^2 ÷ 7.80 m
≈ 18.46 m/s^2 (rounded to two decimal places)
Since the negative acceleration is caused by the kinetic friction between the cars and the road, we can calculate the frictional force using Newton's second law.
Frictional force (f) = mass (m) × acceleration (a)
= (m1 + m2) × a
= (2100 kg + 2780 kg) × 18.46 m/s^2
= 4880 kg × 18.46 m/s^2
≈ 90,014.8 N (rounded to three decimal places)
Finally, we can calculate the coefficient of kinetic friction (μ) using the equation:
Frictional force (f) = coefficient of kinetic friction (μ) × normal force (N)
The normal force (N) is equal to the force of gravity acting on the cars, which can be calculated as:
N = (m1 + m2) × g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = (2100 kg + 2780 kg) × 9.8 m/s^2
= 4880 kg × 9.8 m/s^2
≈ 47824 N (rounded to three decimal places)
Plugging in the values:
90,014.8 N = μ × 47,824 N
Solving for μ:
μ ≈ 1.88 (rounded to two decimal places)
Therefore, the coefficient of kinetic friction between the cars and the road is approximately 1.88.