Question

Two cars are involved in a head-on collision in an alley. The driver of a car with mass, ma = 1,200 kg claims to have been traveling west at via = 7 m/s. The driver of the other car with mass, mb = 1,000 kg claims to have been traveling east at vib = 10 m/s. The investigating officer found that both cars were stuck together and traveled with a common final velocity after the collision.

Which direction would they be moving after the collision?


Question 3 options:

west


east


north


no direction/ not moving

Answer 1

To determine the direction the two cars would be moving after the collision, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.

First, let's define the momentum of each car:

- For car A (mass \( m_a = 1200 \, \text{kg} \) traveling west at \( v_{ia} = 7 \, \text{m/s} \)):
- The momentum of car A is \( p_a = m_a \cdot v_{ia} = 1200 \, \text{kg} \cdot 7 \, \text{m/s} = 8400 \, \text{kg m/s} \) (westward).

- For car B (mass \( m_b = 1000 \, \text{kg} \) traveling east at \( v_{ib} = 10 \, \text{m/s} \)):
- The momentum of car B is \( p_b = m_b \cdot v_{ib} = 1000 \, \text{kg} \cdot 10 \, \text{m/s} = 10000 \, \text{kg m/s} \) (eastward).

Now we need to consider the direction. We'll take west as a negative direction and east as positive.

The total momentum before the collision is:
\[
p_{\text{total}} = p_a + p_b = -8400 \, \text{kg m/s} + 10000 \, \text{kg m/s} = 1600 \, \text{kg m/s} \text{ (eastward)}
\]

After the collision, since the cars are stuck together, they will have a combined mass of:
\[
m_{\text{total}} = m_a + m_b = 1200 \, \text{kg} + 1000 \, \text{kg} = 2200 \, \text{kg}
\]

Let \( v_f \) be the final velocity of the combined mass after the collision. The momentum after collision is given by:
\[
p_{\text{after}} = m_{\text{total}} \cdot v_f
\]

Setting the total momentum before the collision equal to the total momentum after the collision:
\[
1600 \, \text{kg m/s} = 2200 \, \text{kg} \cdot v_f
\]

Now solve for \( v_f \):
\[
v_f = \frac{1600 \, \text{kg m/s}}{2200 \, \text{kg}} \approx 0.727 \, \text{m/s}
\]

Since \( v_f \) is positive, it means the final velocity is directed east.

Thus, the direction in which the two cars would be moving after the collision is **east**.