Question

Select the correct answer.
A car with a mass of 1.5 × 103 kilograms is traveling west at a velocity of 22 meters/second. It hits a stationary car with a mass of 9.0 × 102 kilograms. If the collision is inelastic, what is the final direction and approximate velocity of the two cars?

A.
14 meters/second to the west
B.
14 meters/second to the east
C.
22 meters/second to the east
D.
22 meters/second to the west

Answer 1

In an inelastic collision, momentum is conserved. We can calculate the final velocity of both cars after the collision using the principle of conservation of momentum.

**Given:**
- Mass of the first car \( m_1 = 1.5 \times 10^3 \) kg
- Velocity of the first car \( v_1 = 22 \) m/s (west)
- Mass of the second car \( m_2 = 9.0 \times 10^2 \) kg
- Velocity of the second car \( v_2 = 0 \) m/s (stationary)

**Applying conservation of momentum:**
The total momentum before the collision equals the total momentum after the collision.

\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
\]

Substituting the known values:

\[
(1.5 \times 10^3 \text{ kg}) (22 \text{ m/s}) + (9.0 \times 10^2 \text{ kg}) (0 \text{ m/s}) = (1.5 \times 10^3 \text{ kg} + 9.0 \times 10^2 \text{ kg}) v_f
\]

Calculating the left side:

\[
(1.5 \times 10^3 \times 22) + 0 = 33000 \text{ kg m/s}
\]

Calculating the mass on the right side:

\[
(1.5 \times 10^3 + 9.0 \times 10^2) = 2400 \text{ kg}
\]

Setting up the equation:

\[
33000 = 2400 v_f
\]

To find \( v_f \):

\[
v_f = \frac{33000}{2400} \approx 13.75 \text{ m/s}
\]

Since the first car was moving west and the second car was stationary, the final velocity will also be directed west.

Thus, the approximate final velocity of the two cars will be approximately 14 m/s to the west.

**The correct answer is:**
A. 14 meters/second to the west