Problem 29 Two cars \(A\) and \(B\) slide o... [FREE SOLUTION] (2024)

Short Answer

kinetic friction

Kinetic friction occurs when two objects slide against each other. It's a resistive force that opposes motion. In this exercise, both cars experience kinetic friction because their wheels are locked and sliding on the icy road.
The kinetic friction force can be calculated using the formula:
\[ f_k = \text{normal force} \times \text{coefficient of friction} \]
Since the normal force is the weight of the car (mass times gravitational acceleration), we get:\[ f_k = m \times g \times \text{coefficient of friction} \]
For both cars, we substitute the values: for car A, \( f_k = 1100 \times 9.8 \times 0.13 \text{ N} \) and for car B, \( f_k = 1400 \times 9.8 \times 0.13 \text{ N} \)
Understanding kinetic friction is crucial because it helps us determine the deceleration experienced by each car due to this resistive force.

deceleration

Deceleration is the rate at which an object slows down. It is essentially negative acceleration. Here, the deceleration is caused by kinetic friction.
The deceleration can be found using the relation:
\[ a = \frac{f_k}{m} = g \times \text{coefficient of friction} \]
For our scenario, using the coefficient of kinetic friction (0.13) and the acceleration due to gravity (9.8 m/s²), the deceleration is:
\[ a = 9.8 \times 0.13 = 1.274 \text{ m/s}^2 \]
This value applies to both cars since they have the same coefficient of kinetic friction and are affected equally by gravity. The deceleration value informs us how quickly the cars slow down after the collision.

kinematic equations

Kinematic equations relate the motion parameters of an object, including its initial and final velocities, acceleration, time, and displacement.
In our problem, we use the equation:
\[ v^2 = u^2 - 2ad \]
Here, \( v \) is the final velocity (which is 0, since the cars stop), \( u \) is the initial velocity right after the collision, \( a \) is the deceleration, and \( d \) is the distance traveled.
For car A, we have:
\[ 0 = u_A^2 - 2 \times 1.274 \times 8.2 \]
Solving for \( u_A \):
\[ u_A = \text{√}(2 \times 1.274 \times 8.2) = 4.53 \text{ m/s} \]
For car B, we have:
\[ 0 = u_B^2 - 2 \times 1.274 \times 6.1 \]
Solving for \( u_B \):
\[ u_B = \text{√}(2 \times 1.274 \times 6.1) = 3.95 \text{ m/s} \]
Knowing these initial velocities after impact helps us use the principle of conservation of linear momentum to find the pre-impact speed of car B.

linear momentum

Linear momentum is the product of an object's mass and its velocity. It represents the quantity of motion an object has. Momentum is conserved in isolated systems, meaning the total momentum before and after a collision is the same, assuming no external forces.
The conservation of linear momentum is expressed as:
\[ m_A u_{A\text{ initial}} + m_B u_{B\text{ initial}} = m_A u_A + m_B u_B \]
In this problem, car A is at rest before the collision (\(u_{A\text{initial}} = 0\)). So, the equation simplifies to:
\[ m_B u_{B\text{ initial}} = m_A u_A + m_B u_B \]
Solving for the initial velocity of car B before the collision \(u_{B\text{initial}}\), we get:
\[ u_{B\text{initial}} = \frac{m_A \times u_A + m_B \times u_B}{m_B} = \frac{1100 \times 4.53 + 1400 \times 3.95}{1400} = 4.19 \text{ m/s} \]
However, it is important to note that the external force due to kinetic friction affects the momentum calculation, which limits the accuracy of pure conservation of momentum in this scenario.