Question

Two cars are headed in the same direction on the HWY. The trailing car is moving at 16m/s and has a mass of 1,326 kg. The lead car is moving at 13.3 m/s and has a mass of1,206 kg. The trailing car runs into the lead car and bumps it. Afterwards, the trailing carhas a velocity of 10.3 m/s. What is the velocity of the lead car?

Answer 1

We have to use the law of conservation of momentum, which states the following.

$p_{i1}+p_{i2}=p_{f1}+p_{f2}$

Using the definition of momentum (p = mv), we have the following.

$m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}$

Using the given magnitudes, we replace them and we solve for the velocity of the lead car.

$\begin{gathered} 1,326\operatorname{kg}\cdot16(\frac{m}{s})+1,206\operatorname{kg}\cdot13.3(\frac{m}{s})=1,326\operatorname{kg}\cdot10.3(\frac{m}{s})+1,206\operatorname{kg}\cdot v_{f2} \\ 21,216\operatorname{kg}\cdot\frac{m}{s}+16,039.8\operatorname{kg}\cdot\frac{m}{s}=13,657.8\operatorname{kg}\cdot\frac{m}{s}+1,206\operatorname{kg}\cdot v_{f2} \\ 21,216\operatorname{kg}\cdot\frac{m}{s}+16,039.8\operatorname{kg}\cdot\frac{m}{s}-13,657.8\operatorname{kg}\cdot\frac{m}{s}=1,206\operatorname{kg}\cdot v_{f2} \\ v_{f2}=\frac{23,598\operatorname{kg}\cdot\frac{m}{s}}{1,206\operatorname{kg}} \\ v_{f2}\approx19.57(\frac{m}{s}) \end{gathered}$

Therefore, the velocity of the lead car after the collision is 19.57 meters per second.