Question 6.3: An SUV Versus a Compact Goal Apply conservation of momentum ...
An SUV Versus a Compact Goal Apply conservation of momentum to a one-dimensional inelastic collision.
Problem An SUV with mass 1.80 × 10³ kg is traveling eastbound at +15.0 m/s, while a compact car with mass 9.00 × 10² kg is traveling westbound at -15.0 m/s. (See Fig. 6.9.) The cars collide head-on, becoming entangled. (a) Find the speed of the entangled cars after the collision. (b) Find the change in the velocity of each car. (c) Find the change in the kinetic energy of the system consisting of both cars.
Strategy The total momentum of the cars before the collision, p_{i} , equals the total momentum of the cars after the collision, p_{f} , if we ignore friction and assume the two cars form an isolated system. (This is called the “impulse approximation.”) Solve the momentum conservation equation for the final velocity of the entangled cars. Once the velocities are in hand, the other parts can be solved by substitution.
(a) Find the final speed after collision.
Let m_{1} \text { and } v_{1 i} represent the mass and initial velocity of the SUV, while m_{2} \text { and } v_{2 i} pertain to the compact. Apply conservation of momentum:
\begin{aligned}p_{i} &=p_{f} \\m_{1} v_{1 i}+m_{2} v_{2 i} &=\left(m_{1}+m_{2}\right) v_{f}\end{aligned}
Substitute the values and solve for the final velocity, v_{f} :
\begin{aligned}&\left(1.80 \times 10^{3} kg \right)(15.0 m / s )+\left(9.00 \times 10^{2} kg \right)(-15.0 m / s ) \\&\quad=\left(1.80 \times 10^{3} kg +9.00 \times 10^{2} kg \right) v_{f} \\&v_{f}=5.00 m / s\end{aligned}
(b) Find the change in velocity for each car.
Change in velocity of the SUV:
\Delta v_{1}=v_{f}-v_{1 i}=5.00 m / s -15.0 m / s =-10.0 m / s
Change in velocity of the compact car:
\Delta v_{2}=v_{f}-v_{2 i}=5.00 m / s -(-15.0 m / s )=20.0 m / s
(c) Find the change in kinetic energy of the system.
Calculate the initial kinetic energy of the system:
\begin{aligned}K E_{i}=& \frac{1}{2} m_{1} v_{1 i}{ }^{2}+\frac{1}{2} m_{2} v_{2 i}{ }^{2}=\frac{1}{2}\left(1.80 \times 10^{3} kg \right)(15.0 m / s )^{2} \\&+\frac{1}{2}\left(9.00 \times 10^{2} kg \right)(-15.0 m / s )^{2} \\=& 3.04 \times 10^{5} J\end{aligned}
Calculate the final kinetic energy of the system and the change in kinetic energy, ΔKE.
\begin{aligned}K E_{f} &=\frac{1}{2}\left(m_{1}+m_{2}\right) v_{f}^{2} \\&=\frac{1}{2}\left(1.80 \times 10^{3} kg +9.00 \times 10^{2} kg \right)(5.00 m / s )^{2} \\&=3.38 \times 10^{4} J \\\Delta K E &=K E_{f}-K E_{i}=-2.70 \times 10^{5} J\end{aligned}
Remarks During the collision, the system lost almost 90% of its kinetic energy. The change in velocity of the SUV was only 10.0 m/s, compared to twice that for the compact car. This example underscores perhaps the most important safety feature of any car: its mass. Injury is caused by a change in velocity, and the more massive vehicle undergoes a smaller velocity change in a typical accident.