Question

1. a 2.0 kg cart (cart a) is moving rightward at 3.0 m/s on a frictionless track. it collides elastically and head - on with a 1.0 kg cart (cart b) moving leftward at 1.0 m/s. after the collision, cart b moves rightward at 3.67 m/s. determine the final velocity (speed and direction) of cart a. (1 point) select your answer

Explanation:

Step1: Define the principle

Use conservation of momentum \(p = mv\), where \(p\) is momentum, \(m\) is mass and \(v\) is velocity. Let right - ward be the positive direction. Initial momentum of system \(p_i=m_Av_{iA}+m_Bv_{iB}\), and final momentum \(p_f=m_Av_{fA}+m_Bv_{fB}\). According to conservation of momentum \(p_i = p_f\).

Step2: Substitute the given values

Given \(m_A = 2.0\ kg\), \(v_{iA}=3.0\ m/s\), \(m_B = 1.0\ kg\), \(v_{iB}=- 1.0\ m/s\), \(v_{fB}=3.67\ m/s\).
\(m_Av_{iA}+m_Bv_{iB}=m_Av_{fA}+m_Bv_{fB}\)
\(2.0\times3.0+1.0\times(-1.0)=2.0\times v_{fA}+1.0\times3.67\)

Step3: Solve for \(v_{fA}\)

First, simplify the left - hand side of the equation: \(6.0 - 1.0=5.0\).
The equation becomes \(5.0 = 2.0v_{fA}+3.67\).
Subtract 3.67 from both sides: \(2.0v_{fA}=5.0 - 3.67=1.33\).
Then divide both sides by 2.0: \(v_{fA}=\frac{1.33}{2.0}=0.665\ m/s\)

Answer:

The final velocity of Cart A is \(0.665\ m/s\) to the right.