Question

part 3: collision calculation 5 points each (box your answer)
question: car a with momentum of 3000 kg.m/s collided with carb that was parked. after the collision, both cars moved to the left. if car a has a mass of 500kg and carb has a mass of 1000kg, what is the final velocity of both cars? show work.
question: object a has a mass of 2kg colliding with object b with a mass of 1kg. both objects bounce off each other and retain their shape after the collision. the final velocity of object a is 4m/s and the final velocity of object b is 5m/s. if the initial velocity of object a is 5m/s, what is the initial velocity of object b? show work.

Explanation:

First Question

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(p_{initial}=p_{final}\). Initially, car B is parked, so its initial momentum \(p_{B,i} = 0\) (since \(p = mv\) and \(v_{B,i}=0\)). The initial momentum of the system is \(p_{i}=p_{A,i}+p_{B,i}=3000\space kg\cdot m/s+ 0\space kg\cdot m/s = 3000\space kg\cdot m/s\). After the collision, the two - car system has a combined mass \(m = m_{A}+m_{B}\) and a common velocity \(v\). The final momentum \(p_{f}=(m_{A} + m_{B})v\), where \(m_{A}=500\space kg\) and \(m_{B}=1000\space kg\).

Step2: Solve for the final velocity \(v\)

Using \(p_{i}=p_{f}\), we substitute the values into the equation \(3000=(500 + 1000)v\). First, calculate the sum of the masses: \(m_{A}+m_{B}=1500\space kg\). Then, solve for \(v\) using the formula \(v=\frac{p_{i}}{m_{A}+m_{B}}\). So, \(v=\frac{3000}{1500}\space m/s\)

Step1: Apply the law of conservation of momentum

The law of conservation of momentum is \(m_{A}v_{A,i}+m_{B}v_{B,i}=m_{A}v_{A,f}+m_{B}v_{B,f}\). We are given that \(m_{A} = 2\space kg\), \(v_{A,i}=5\space m/s\), \(v_{A,f}=4\space m/s\), \(m_{B}=1\space kg\), and \(v_{B,f}=5\space m/s\).

Step2: Rearrange the formula to solve for \(v_{B,i}\)

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Substitute the values: \(m_{A}(v_{A,f}-v_{A,i})=2\times(4 - 5)=- 2\space kg\cdot m/s\) and \(m_{B}v_{B,f}=1\times5 = 5\space kg\cdot m/s\). Then \(v_{B,i}=\frac{-2 + 5}{1}\space m/s\)

Answer:

\(\boxed{2\space m/s}\)

Second Question