Question

question 3
suppose m1 (1 kg) is moving at 3 m/s in the +x direction. m2 (2 kg) is moving at 1 m/s in the +x direction. if they collide and stick together, what will be their final velocity (in m/s)?
add your answer
integer, decimal, or e notation allowed

question 4
suppose m1 (1 kg) is moving at 3 m/s in the +x direction. m2 (2 kg) is moving at 1 m/s in the +x direction. they collide and stick together. what is the magnitude of the difference in the kinetic energy (in joules) before and after the collision? use magnitude only.
add your answer
integer, decimal, or e notation allowed

Explanation:

Question 3

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=(m_1 + m_2)v_f\). Here, \(m_1 = 1\space kg\), \(v_1=3\space m/s\), \(m_2 = 2\space kg\), \(v_2 = 1\space m/s\).
Substitute the values into the formula: \((1\times3)+(2\times1)=(1 + 2)v_f\)

Step2: Solve for \(v_f\)

First, calculate the left - hand side: \(3+2 = 5\). Then the equation becomes \(5=3v_f\). Solving for \(v_f\), we get \(v_f=\frac{5}{3}\approx1.67\space m/s\)

Step1: Calculate the initial kinetic energy

The formula for kinetic energy is \(K=\frac{1}{2}mv^{2}\).
For \(m_1 = 1\space kg\) and \(v_1 = 3\space m/s\), \(K_1=\frac{1}{2}\times1\times3^{2}=\frac{9}{2}=4.5\space J\)
For \(m_2 = 2\space kg\) and \(v_2 = 1\space m/s\), \(K_2=\frac{1}{2}\times2\times1^{2}=1\space J\)
The initial kinetic energy \(K_i=K_1 + K_2=4.5 + 1=5.5\space J\)

Step2: Calculate the final kinetic energy

We know from Question 3 that \(v_f=\frac{5}{3}\space m/s\) and \(m=m_1 + m_2=3\space kg\)
Using \(K=\frac{1}{2}mv^{2}\), \(K_f=\frac{1}{2}\times3\times(\frac{5}{3})^{2}=\frac{1}{2}\times3\times\frac{25}{9}=\frac{25}{6}\approx4.17\space J\)

Step3: Calculate the difference in kinetic energy

\(\Delta K=K_i - K_f\)
\(\Delta K=5.5-\frac{25}{6}=\frac{33 - 25}{6}=\frac{8}{6}=\frac{4}{3}\approx1.33\space J\)

Answer:

\(1.67\)

Question 4