Question

in a lab experiment, a student is trying to apply the conservation of momentum. two identical balls, each with a mass of 1.0 kg, roll toward each other and collide. the velocity is measured before and after each collision. the collected data is shown below.

	initial velocity ball a (m/s)	initial velocity ball b (m/s)	final velocity ball a (m/s)	final velocity ball b (m/s)
trial 2	+0.5	-1.5	-0.5	-0.5
trial 3	+2.0	+1.0	+1.0	-2.0
trial 4	+0.5	-1.0	+1.5	-1.5

which trial shows the conservation of momentum in a closed system?
○ trial 1
○ trial 2
○ trial 3
○ trial 4

Explanation:

1. First, recall the formula for momentum:

• The momentum of an object is \(p = mv\), and for a two - object system, the initial total momentum \(p_{i}=m_{A}v_{iA}+m_{B}v_{iB}\) and the final total momentum \(p_{f}=m_{A}v_{fA}+m_{B}v_{fB}\). Since \(m_{A}=m_{B} = 1.0\ kg\), we can simplify the momentum - conservation check to \(v_{iA}+v_{iB}=v_{fA}+v_{fB}\).

1. Analyze each trial:

• Trial 1:
• Initial velocities: \(v_{iA}= + 1.0\ m/s\), \(v_{iB}=-2.0\ m/s\). So, \(v_{iA}+v_{iB}=1.0+( - 2.0)=-1.0\ m/s\).
• Final velocities: \(v_{fA}=-2.0\ m/s\), \(v_{fB}=-1.0\ m/s\). So, \(v_{fA}+v_{fB}=-2.0+( - 1.0)=-3.0\ m/s\). Since \(v_{iA}+v_{iB}

eq v_{fA}+v_{fB}\), momentum is not conserved in this trial.

• Trial 2:
• Initial velocities: \(v_{iA}= + 0.5\ m/s\), \(v_{iB}=-1.5\ m/s\). So, \(v_{iA}+v_{iB}=0.5+( - 1.5)=-1.0\ m/s\).
• Final velocities: \(v_{fA}=-0.5\ m/s\), \(v_{fB}=-0.5\ m/s\). So, \(v_{fA}+v_{fB}=-0.5+( - 0.5)=-1.0\ m/s\). Since \(v_{iA}+v_{iB}=v_{fA}+v_{fB}\), momentum is conserved in this trial.
• Trial 3:
• Initial velocities: \(v_{iA}= + 2.0\ m/s\), \(v_{iB}=+1.0\ m/s\). So, \(v_{iA}+v_{iB}=2.0 + 1.0=3.0\ m/s\).
• Final velocities: \(v_{fA}=+1.0\ m/s\), \(v_{fB}=-2.0\ m/s\). So, \(v_{fA}+v_{fB}=1.0+( - 2.0)=-1.0\ m/s\). Since \(v_{iA}+v_{iB}

eq v_{fA}+v_{fB}\), momentum is not conserved in this trial.

• Trial 4:
• Initial velocities: \(v_{iA}= + 0.5\ m/s\), \(v_{iB}=-1.0\ m/s\). So, \(v_{iA}+v_{iB}=0.5+( - 1.0)=-0.5\ m/s\).
• Final velocities: \(v_{fA}=+1.5\ m/s\), \(v_{fB}=-1.5\ m/s\). So, \(v_{fA}+v_{fB}=1.5+( - 1.5)=0\ m/s\). Since \(v_{iA}+v_{iB}

eq v_{fA}+v_{fB}\), momentum is not conserved in this trial.

Answer:

Trial 2