Problem 36 - University Physics with Modern Physics

Question

A \(1050\) -kg sports car is moving westbound at 15.0 \(\mathrm{m} / \mathrm{s}\) on a level road when it collides with a 6320 -kg truck driving east on the same road at 10.0 \(\mathrm{m} / \mathrm{s}\) . The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that it and the car are both stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

Step-by-Step Solution

Answer

(a) -6.43 m/s east; (b) -2.49 m/s; (c) Greater change in scenario (b).

Step 1: Determine the Initial Momentum

First, calculate the momentum of each vehicle before the collision. The momentum of an object is given by the equation \( p = m \cdot v \).For the car: \( p_{\text{car}} = 1050 \times 15.0 = 15750 \, \text{kg} \cdot \text{m/s} \) (westbound).For the truck: \( p_{\text{truck}} = 6320 \times (-10.0) = -63200 \, \text{kg} \cdot \text{m/s} \) (eastbound, so the velocity is negative).The total initial momentum is: \( p_{\text{total}} = 15750 - 63200 = -47450 \, \text{kg} \cdot \text{m/s} \).

Step 2: Apply Conservation of Momentum for Collision

Since the collision is perfectly inelastic, we use the conservation of momentum to find the final velocity.The total momentum before collision equals the total momentum after collision:\[ m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot v_{\text{truck}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_{\text{final}} \]Plugging in the values:\[ 15750 - 63200 = (1050 + 6320) \cdot v_{\text{final}} \]\[ -47450 = 7370 \cdot v_{\text{final}} \]Solve for \( v_{\text{final}} \):\[ v_{\text{final}} = \frac{-47450}{7370} \approx -6.43 \, \text{m/s} \]The negative direction indicates that the combined vehicles move east after the collision.

Step 3: Determine the Truck's Speed to Stop Both Vehicles

To stop both vehicles after the collision, the final velocity \( v_{\text{final}} = 0 \). Re-apply the conservation of momentum:\[ m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot v_{\text{truck}} = 0 \]Solve for \( v_{\text{truck}} \):\[ 15750 + 6320 \cdot v_{\text{truck}} = 0 \]\[ v_{\text{truck}} = -\frac{15750}{6320} \approx -2.49 \, \text{m/s} \]

Step 4: Calculate Change in Kinetic Energy for Both Situations

Kinetic energy is given by: \( KE = \frac{1}{2} m v^2 \).**Situation (a):**Initial kinetic energy:\[ KE_{\text{initial}} = \frac{1}{2} (1050 \times 15^2) + \frac{1}{2} (6320 \times 10^2) \approx 118125 + 316000 = 434125 \, \text{J} \]Final kinetic energy:\[ KE_{\text{final}} = \frac{1}{2} (7370 \times (-6.43)^2) \approx 151826 \, \text{J} \]Change in kinetic energy:\[ \Delta KE_{a} = 151826 - 434125 = -282299 \, \text{J} \]**Situation (b):**Since the final velocity is zero, the final kinetic energy is zero, so the change in kinetic energy is simply:\[ \Delta KE_{b} = 0 - 434125 = -434125 \, \text{J} \]In magnitude, the change is greater in scenario (b).

Key Concepts

Conservation of Momentum

The conservation of momentum is a fundamental principle in physics that applies to all collisions. It states that the total momentum of a closed system remains constant provided no external forces act on it. In our scenario, the sports car and the truck form a closed system during their collision.

To apply this principle, consider the total momentum before the collision. You'll first calculate the momentum for each involved object, using this formula:

• Momentum of the car: \( p = m \cdot v \)

With the momentum for both the sports car and truck calculated, their sum gives the total initial momentum.

• For the car moving west, with mass \( 1050 \) kg and velocity \( 15.0 \text{ m/s} \):
  \( p_{\text{car}} = 1050 \times 15 = 15750 \text{ kg} \cdot \text{m/s} \)
• For the truck moving east, with mass \( 6320 \) kg and velocity \(-10.0 \text{ m/s} \):
  \( p_{\text{truck}} = 6320 \times (-10) = -63200 \text{ kg} \cdot \text{m/s} \)
• Total initial momentum: \( p_{\text{total}} = 15750 - 63200 = -47450 \text{ kg} \cdot \text{m/s} \)

The negative sign indicates that the system’s momentum is directed eastward. After the collision, this momentum ideally remains unchanged, aiding in finding the new velocity when both vehicles move as one unit.

Kinetic Energy Change

Kinetic energy is the energy possessed by an object due to its motion, and it is calculated using the formula: \( KE = \frac{1}{2} m v^2 \). In inelastic collisions, such as the one in our scenario, some kinetic energy is transformed into other forms of energy, like sound or heat, causing a change in the total kinetic energy.

For our scenarios:

• **Initial kinetic energy** is the sum of the kinetic energies of the car and truck before impact:
  \( KE_{\text{initial}} = \frac{1}{2} (1050 \cdot 15^2) + \frac{1}{2} (6320 \cdot 10^2) = 434125 \text{ J} \).
• **Final kinetic energy** for scenario (a) is calculated from the combined mass moving at the final velocity:
  \( KE_{\text{final}} = \frac{1}{2} (7370 \cdot 6.43^2) = 151826 \text{ J} \).
• The **change in kinetic energy** is simply the final minus initial, indicating energy loss during collision:
  \( \Delta KE_{a} = 151826 - 434125 = -282299 \text{ J} \).

In scenario (b), where the final speed is zero, the entire initial kinetic energy is lost, leading to \( \Delta KE_{b} = -434125 \text{ J} \). Clearly, more energy is lost in scenario (b).

Velocity After Collision

Post-collision velocity can be deduced using the principle of momentum conservation, essential for understanding inelastic collisions. After the vehicles collide and lock together, they move with a common velocity.

The conservation of momentum formula states:

• Total momentum before the collision = total momentum after the collision:
  \( 1050 \cdot 15 - 6320 \cdot 10 = (1050 + 6320) \cdot v_{\text{final}} \)
• Solving the equation gives:
  \( v_{\text{final}} = \frac{-47450}{7370} \approx -6.43 \text{ m/s} \)

The negative velocity indicates the direction of movement is eastward. Understanding the velocity change helps in analyzing accidents and their impact sizes.

Stopping Condition in Collisions

The stopping condition for both vehicles in a collision implies bringing their common post-collision velocity to zero. This condition allows us to determine the necessary speed of the truck to completely stop both vehicles in their tracks.

Given the equation for momentum conservation:

• Final velocity \( v_{\text{final}} = 0 \), simplifies the conservation equation to:
  \( 1050 \cdot 15 + 6320 \cdot v_{\text{truck}} = 0 \)
• Solving gives the required truck speed:
  \( v_{\text{truck}} = -\frac{15750}{6320} \approx -2.49 \text{ m/s} \)

This calculation shows the importance of initial velocities in determining whether vehicles can stop after an impact. Further implications relate to the severity of accidents and energy transfers during such events.