Question

A 2200 kg car traveling at 8.7 m/s collides with a 2750 kg car that is initially at rest at the stoplight. The cars stick together and move 2.00 m before friction causes them to stop.

Determine the coefficient of kinetic friction between 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.

Answer 1

To solve this problem we will apply three conceps that will take us to the coefficient of dynamic friction. We will start from the momentum conservation theorem to find the final velocity of the bodies. From this speed we will use the linear motion kinematic equations to find the acceleration.

Finally by equilibrium of Forces, the friction force must be equal to the force caused by the acceleration in the mass of the bodies. Let's start

By Conservation of momentum

$m_1v_1+m_1v_2=(m_1+m_2)vf$

$2200*8.7+2750*0=(2200+2750)v_f$

$v_f=3.86 m/s$

So combined cars velocity after impact = 3.86 m/s

Applying the cinematic equations we have that the acceleration is

$v_f^2=v_0^2 + 2ax$

Replacing,

$3.86^2=0+2a(2)$

$a = 3.72m/s^2$

Now by equilibrium in force we have that,

$F_f=ma$

$\mu N = ma$

$\mu (mg) = ma$

$\mu = \frac{a}{g}$

$\mu = \frac{3.72}{9.8}$

$\mu = 0.379$

Therefore the coefficient of kinetic friction between the cars and the road is 0.379

Answer 2

The coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant is 0.38

How to determine the common velocity

• Mass of car 1 (m₁) = 2200 Kg
• Velocity of cat 1 (v₁) = 8.7 m/s
• Mass of car 2 (m₂) = 2750 Kg
• Velocity of car 2 (v₂) = 0 m/s
• Common velocity (v) =?

v = (m₁v₁ + m₂v₂) / (m₁ + m₂)

v = [(2200 × 8.7) + (2750 × 0)] / (2200 + 2750)

v = 19140 / 4950

v = 3.87 m/s

How to determine the deceleration

• Initial velocity (u) = 3.87 m/s
• Final velocity (v) = 0 m/s
• Distance (s) = 2 m
• Deceleration (a) =?

v² = u² + 2as

0² = 3.87² + (2 × a × 2)

0 = 14.9769 + 4a

Collect like terms

4a = 0 – 14.9769

4a = –14.9769

Divide both side by 4

a = –14.9769 / 4

a = –3.74 m/s²

How to determine the coefficient of friction

• Acceleration (a) = 3.74 m/s²
• Acceleration due to gravity (g) = 9.8 m/s²

• Coefficient of friction (μ) =?

Frictional force (N) = coefficient of friction (μ) × normal reaction (N)

F = μN

But

F = ma

N = mg

ma = μmg

Cancle out m

a = μg

Divide both side by g

μ = a / g

μ = 3.74 / 9.8

μ = 0.38

Learn more about momentum:

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