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Chapter 7: Problem 29
Two cars \(A\) and \(B\) slide on an icy road as they attempt to stop at a trafficlight. The mass of \(A\) is \(1100 \mathrm{~kg}\), and the mass of \(B\) is \(1400\mathrm{~kg}\). The coefficient of kinetic friction between the locked wheelsof either car and the road is 0.13. Car \(A\) succeeds in stopping at the light,but car \(B\) cannot stop and rear-ends car \(A\). After the collision, \(A\) stops\(8.2 \mathrm{~m}\) ahead of its position at impact, and \(B 6.1 \mathrm{~m}\)ahead; see Fig. 7-27. Both drivers had their brakes locked throughout theincident. Using the material in Chapters 2 and 6 , find the speed of (a) car\(A\) and (b) car \(B\) immediately after impact. (c) Use conservation oftranslational momentum to find the speed at which car \(B\) struck car \(A\). Onwhat grounds can the use of momentum conservation be criticized here?
Short Answer
Expert verified
Car A's speed after impact is 4.53 m/s, car B's speed is 3.95 m/s, and car B's speed before impact is 4.19 m/s. Conservation of momentum may be criticized due to external frictional force.
Step by step solution
01
- Understand the problem
We need to find the speeds of cars A and B after the collision and then use the conservation of momentum to find the speed of B before the collision. We will use the given masses and the distances traveled after impact to calculate the initial velocities using kinematics and friction.
02
- Use kinetic friction to find deceleration
The frictional force acting on each car is given by: \[ f_k = \text{normal force} \times \text{coefficient of friction} = m \times g \times \text{coefficient of friction} onumber \] This force causes deceleration: \[ f_k = m \times a \rightarrow a = \frac{f_k}{m} = g \times \text{coefficient of friction} onumber \] Hence, the deceleration for both cars is: \[ a = 9.8 \times 0.13 = 1.274 \text{ m/s}^2 onumber \]
03
- Apply kinematic equations to find initial velocities after impact
Using the equation \[ v^2 = u^2 - 2ad onumber \] where v (final velocity) is 0 since the cars stop, we can solve for u (initial velocity immediately after impact): For car A: \[ 0 = u_A^2 - 2 \times 1.274 \times 8.2 \] Using algebra: \[ u_A = \sqrt{2 \times 1.274 \times 8.2} = 4.53 \text{ m/s} onumber \] For car B: \[ 0 = u_B^2 - 2 \times 1.274 \times 6.1 \] Using algebra: \[ u_B = \sqrt{2 \times 1.274 \times 6.1} = 3.95 \text{ m/s} onumber \]
04
- Apply conservation of momentum to find initial velocity of car B before impact
Using the conservation of linear momentum: \[ m_A \times u_{A\text{ initial}} + m_B \times u_{B\text{ initial}} = m_A \times u_A + m_B \times u_B onumber \] Since car A was at rest before the collision: \[ 0 + m_B \times u_{B\text{ initial}} = m_A \times 4.53 + m_B \times 3.95 onumber \] Solving for the initial velocity of car B ( u_{B\text{ initial}} ): \[ u_{B\text{ initial}} = \frac{m_A \times u_A + m_B \times u_B}{m_B} = \frac{1100 \times 4.53 + 1400 \times 3.95}{1400} onumber \] Using algebra: \[ u_{B\text{ initial}} = 4.19 \text{ m/s} onumber \]
05
- Critique the use of momentum conservation
The use of momentum conservation can be criticized because the frictional force exerted by the road is an external force on the system (cars A and B). In a collision involving external forces, conservation of momentum might not hold strictly since these external forces can cause additional changes in momentum.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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