Question

1. a 25 kg turkey is fired from a 1.1 x 10³ kg turkey launcher. if the horizontal velocity of the turkey is 325 m/s east, what is the recoil velocity of the launcher? (7.4 m/s west)

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=0\) (since the initial momentum of the system (turkey + launcher) is \(0\) as it is at rest initially). Let \(m_1 = 25\space kg\), \(v_1=325\space m/s\) (east - we can take east as positive), \(m_2 = 1.1\times10^{3}\space kg\) and \(v_2\) is the recoil velocity of the launcher.
So, \(m_1v_1+m_2v_2 = 0\).

Step2: Solve for \(v_2\)

Rearrange the formula \(v_2=-\frac{m_1v_1}{m_2}\).
Substitute \(m_1 = 25\space kg\), \(v_1 = 325\space m/s\), \(m_2=1.1\times 10^{3}\space kg\) into the formula:
\(v_2=-\frac{25\times325}{1.1\times 10^{3}}\)
\(v_2=-\frac{8125}{1100}\approx - 7.4\space m/s\)

Answer:

The recoil velocity of the launcher is approximately \(7.4\space m/s\) west.