Question

3.3.2 test (cst): forces
question 20 of 20
a water - balloon launcher with a mass of 2.2 kg is suspended on a wire. it fires a 0.85 kg balloon to the north at a velocity of 13.0 m/s. what is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?
a. 5.5 m/s north
b. 5.0 m/s south
c. 33.6 m/s south
d. 6.3 m/s south

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=m_1v_1'+m_2v_2'\). Initially, the total momentum of the system (launcher + balloon) is \(0\) (since both are at rest, \(v_1 = v_2=0\)). So, \(0 = m_{balloon}v_{balloon}+m_{launcher}v_{launcher}\).

Step2: Solve for the velocity of the launcher

We can rewrite the equation from Step 1 as \(v_{launcher}=-\frac{m_{balloon}v_{balloon}}{m_{launcher}}\).
Substitute \(m_{balloon} = 0.85\space kg\), \(v_{balloon}=13.0\space m/s\), and \(m_{launcher}=2.2\space kg\) into the formula:
\(v_{launcher}=-\frac{0.85\times13.0}{2.2}\)
\(v_{launcher}=-\frac{11.05}{2.2}\approx - 5.0\space m/s\)
The negative sign indicates the direction is opposite to the balloon's direction (south, since the balloon moves north).

Answer:

B. \(5.0\space m/s\) south