Question

In an elastic collision involving known masses and initial momenta, how many unknown quantities are there after the collision? How many equations are there? In a sticking collision involving known masses and initial momenta, how many unknown quantities are there after the collision? Explain how you can determine the amount of kinetic energy change.

Short answer

After the collision, there are three equations and two unknown quantities. Internal energy increases when kinetic energy decreases.

Step-by-step solution

Given variable

There are only two unknowns after the impact for sticking this collision: each mass's final speed and direction.

The amount of decrease in kinetic energy (the difference between before and after the collision) is equal to the amount of gain in internal energy in this collision.

The concept of elastic collision

If the total system kinetic energy before the collision equals the energy after the collision, then it is an elastic collision.

To determine the amount of kinetic energy change

Consider an elastic collision with known masses and beginning momenta; because the first momenta are known, the initial velocities are likewise known, because momentum is the product of mass and velocity. Because we have an elastic collision and three equations (two for the x and y components of momentum and one for energy), the laws of conservation of momentum and energy hold true after the collision. The unknown numbers after the collision are the final velocities of the two masses.

Suppose a head-on elastic collision occurs between two trolleys on a track. We want to know the final velocities (subscript f) for both the trolleys, but are only given the initial velocities$v_{Ai}$and$v_{Bi}$. Applying conservation of momentum we can see that we have one equation with two unknowns,$v_{Af}$and$v_{Bf}$:

$m_A{v}_{Ai}+m_B{v}_{Bi}=m_A{v}_{Af}+m_B{v}_{Bf}$

Because kinetic energy is also conserved, we simultaneously have another constraint:

$\frac{1}{2}{m}_A{v}_{Ai}^2+\frac{1}{2}{m}_B{v}_{Bi}^2=\frac{1}{2}{m}_A{v}_{Af}^2+\frac{1}{2}{m}_B{v}_{Bf}^2$

Solving these equations is somewhat tedious. For now, we simply state the result:

$v_{Af}=\left(\frac{m_A-m_B}{m_A+m_B}\right)v_{Ai}+\left(\frac{2m_B}{m_A+m_B}\right)v_{Bi}$

$v_{Bf}=\left(\frac{2m_A}{m_A+m_B}\right)v_{Ai}+\left(\frac{m_B-m_A}{m_A+m_B}\right)v_{Bi}$

As we now have two equations with two unknowns, we know that we can completely solve the system using simultaneous equations to determine both velocities.

For sticking this collision there are only two unknowns after the collision, each mass's final speed and direction.

In this collision the amount of decrease in kinetic energy (difference between before and after the collision takes place) is the amount of increase in internal energy

There are three equations and two unknown quantities after the collision.

Decrease in kinetic energy $=$increase in internal energy.