91Ó°ÊÓ

During an inelastic collision, two objects crash into each other and stick together. This implies that they move as a single combined object after the collision. Contrary to elastic collisions, where kinetic energy is conserved, in inelastic collisions, only momentum is conserved. Energy transformation into other forms, like heat or deformation, often occurs.
In the exercise we have on hand, when the 5-Mg bus and 2-Mg car collide, they become entangled. As they stick together, this is a textbook example of an inelastic collision. The focus shifts from merely the speed to how the system behaves as a whole. By considering only the momentum, we can bypass the complexities introduced by various energy conversions post-collision.

Momentum is a fundamental concept in physics utilized to predict the outcome of interactions like collisions. It is given by the product of mass and velocity and is expressed mathematically as: \[ p = m \times v \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In this exercise, calculating the individual momenta of the bus and car before the collision is necessary.

• The bus, with a mass of 5 Mg moving at 20 m/s, has a momentum of \( 5 \times 20 = 100 \, \text{Mg m/s} \).
• The car, with a mass of 2 Mg moving at 15 m/s, has a momentum of \( 2 \times 15 = 30 \, \text{Mg m/s} \).

Adding these values gives us the total momentum of the system just before the collision, which is crucial because we utilize this in our conservation of momentum calculations.

To find the common velocity of two entangled objects after an inelastic collision, we apply the principle of conservation of momentum. According to this principle, the total momentum before collision equals the total momentum after collision. This can be represented with the equation: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) V \] Given the exercise, the total initial momentum we calculated as \( 100 \, \text{Mg m/s} + 30 \, \text{Mg m/s} \) must equate to the total final momentum \( V \cdot (5 \text{ Mg} + 2 \text{ Mg}) \).

• Plugging in the known values, our equation becomes \( 130 \, \text{Mg m/s} = 7 V \).
• Solving for \( V \), the common velocity is determined to be \( \frac{130}{7} \, \text{m/s} \).

This calculation shows the new shared velocity post-collision, wrapping up our problem solution.