for a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass…

Explanation:

Step1: Recall conservation of momentum

In a closed - system (no external forces), the total momentum before the collision $p_i$ is equal to the total momentum after the collision $p_f$. The initial momentum of the system is $p_i = m_{truck}v_{truck}+m_{car}v_{car}$, where $m_{truck}=2000\ kg$, $v_{truck} = 10\ m/s$, $m_{car}=1000\ kg$ and $v_{car}=0\ m/s$. So $p_i=2000\times10+1000\times0 = 20000\ kg\cdot m/s$. After the collision, the two objects stick together, so $m_{total}=m_{truck}+m_{car}=2000 + 1000=3000\ kg$. Let the final velocity be $v_f$. By conservation of momentum $p_f=p_i$, so $m_{total}v_f=p_i$, and $v_f=\frac{p_i}{m_{total}}=\frac{20000}{3000}=\frac{20}{3}\approx6.67\ m/s$.

Step2: Analyze elastic and in - elastic collisions

Since the truck and car stick together, this is an in - elastic collision. In an in - elastic collision, kinetic energy is not conserved.

Step3: Check momentum and speed statements

The momentum before and after the collision is the same due to conservation of momentum. The initial speed of the truck is $v_{truck}=10\ m/s$ and the final speed of the combined vehicles is $v_f=\frac{20}{3}\ m/s\lt10\ m/s$.

Answer:

The speed of the combined vehicles is less than the initial speed of the truck.