Question

1. a 1.50 x 10³ kg car traveling at 100 km/h south collides with a 1.20 x 10³ kg car traveling north at 100 km/h. the heavier car continues to move south after the collision, but slows to 25.0 km/h. how fast is the lighter car moving after the collision?

Explanation:

Step1: Convert velocities to SI units

Let's take South as positive.
Initial velocity of heavier car \(v_{1i}=100\ km/h=\frac{100\times1000}{3600}\ m/s\approx27.78\ m/s\)
Initial velocity of lighter car \(v_{2i}=- 100\ km/h=-\frac{100\times1000}{3600}\ m/s\approx - 27.78\ m/s\)
Final velocity of heavier car \(v_{1f}=25\ km/h=\frac{25\times1000}{3600}\ m/s\approx6.94\ m/s\)

Step2: Apply conservation of momentum

The formula for conservation of momentum is \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\)
where \(m_1 = 1.50\times10^{3}\ kg\), \(m_2=1.20\times 10^{3}\ kg\)
Substitute the values:
\((1.50\times10^{3})\times27.78+(1.20\times10^{3})\times(- 27.78)=(1.50\times10^{3})\times6.94+(1.20\times10^{3})\times v_{2f}\)

Step3: Simplify the equation

First, expand the left - hand side:
\(1.50\times10^{3}\times27.78-1.20\times10^{3}\times27.78=(1.50\times6.94 + 1.20v_{2f})\times10^{3}\)
\((1.50 - 1.20)\times27.78\times10^{3}=(10.41+1.20v_{2f})\times10^{3}\)
\(0.3\times27.78=10.41 + 1.20v_{2f}\)

Step4: Solve for \(v_{2f}\)

\(8.334=10.41+1.20v_{2f}\)
\(1.20v_{2f}=8.334 - 10.41\)
\(1.20v_{2f}=- 2.076\)
\(v_{2f}=\frac{-2.076}{1.20}=- 1.73\ m/s\)

Convert back to \(km/h\): \(v_{2f}=-1.73\times\frac{3600}{1000}\ km/h=- 6.23\ km/h\)

Answer:

The lighter car is moving at approximately \(6.23\ km/h\) North.