Momentum and collisions

Example 1: impulse on a baseball

Problem A 0.145 kg baseball comes toward a bat at 38.0 m/s and leaves in the opposite direction at 46.0 m/s. The contact lasts 1.20 ms. Find the impulse and the average force.

1. Take the direction the ball leaves as positive, so $v_1 = -38.0\ \text{m/s}$ and $v_2 = +46.0\ \text{m/s}$.
2. Change in momentum.
   $$\Delta p = m(v_2 - v_1) = (0.145\ \text{kg})(46.0 - (-38.0))\ \text{m/s} = 12.18\ \text{kg·m/s}$$
3. Average force. Convert 1.20 ms to 0.00120 s.
   $$F = \frac{\Delta p}{\Delta t} = \frac{12.18\ \text{kg·m/s}}{0.00120\ \text{s}} = 10{,}150\ \text{N}$$

Answer Impulse 12.2 N·s, average force $1.02 \times 10^4$ N

Example 2: cars that stick together

Problem A 1500 kg car moving at 20.0 m/s hits a stopped 1000 kg car, and they lock together. How fast do they move just after, and how much kinetic energy is lost?

1. Momentum before: only the first car moves.
   $$p = (1500\ \text{kg})(20.0\ \text{m/s}) + 0 = 30{,}000\ \text{kg·m/s}$$
2. After, both move together with mass 2500 kg.
   $$v' = \frac{30{,}000\ \text{kg·m/s}}{2500\ \text{kg}} = 12.0\ \text{m/s}$$
3. Kinetic energy before and after.
   $$KE_1 = \tfrac{1}{2}(1500)(20.0)^2 = 300{,}000\ \text{J} \qquad KE_2 = \tfrac{1}{2}(2500)(12.0)^2 = 180{,}000\ \text{J}$$
4. Lost to crumpling, heat and sound.
   $$300{,}000 - 180{,}000 = 120{,}000\ \text{J}$$

Answer 12.0 m/s, with $1.20 \times 10^5$ J of kinetic energy lost

Example 3: pushing apart (recoil)

Problem Two skaters stand still on ice. A 60.0 kg skater and a 40.0 kg skater push off each other. The 40.0 kg skater moves away at 3.00 m/s. How fast does the other skater move?

1. Total momentum before is zero, so it is still zero after.
   $$0 = (40.0)(3.00) + (60.0)v'$$
2. Solve.
   $$v' = -\frac{120\ \text{kg·m/s}}{60.0\ \text{kg}} = -2.00\ \text{m/s}$$
3. The minus sign means the opposite direction.

Answer 2.00 m/s in the opposite direction

Example 4: an elastic collision of equal masses

Problem A ball moving at 3.0 m/s hits an identical ball at rest head-on, in a perfectly elastic collision. What are their velocities afterward?

1. For an elastic head-on collision with the second object at rest, the standard results are:
   $$v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1 \qquad v_2' = \frac{2m_1}{m_1 + m_2} v_1$$
2. Equal masses make $m_1 - m_2 = 0$ and $\tfrac{2m_1}{m_1 + m_2} = 1$.
   $$v_1' = 0 \qquad v_2' = v_1 = 3.0\ \text{m/s}$$
3. Check: momentum $m(3.0)$ before and after, and kinetic energy $\tfrac{1}{2}m(3.0)^2$ before and after.

Answer The first ball stops and the second moves off at 3.0 m/s