Question

1. a 25 kg turkey is fired from a 1100 kg turkey launcher. if the horizontal velocity of the turkey is 325 m/s east, what is the recoil of the launcher?

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(p_{initial}=p_{final}\). Initially, the total momentum \(p_{initial} = 0\) (both the turkey and the launcher are at rest). Let \(m_1 = 25\space kg\), \(v_1=325\space m/s\), \(m_2 = 1100\space kg\) and \(v_2\) be the velocity of the launcher. Then \(0=m_1v_1 + m_2v_2\).

Step2: Solve for \(v_2\)

Rearrange the equation \(0=m_1v_1 + m_2v_2\) to get \(v_2=-\frac{m_1v_1}{m_2}\). Substitute \(m_1 = 25\space kg\), \(v_1 = 325\space m/s\) and \(m_2=1100\space kg\) into the formula: \(v_2=-\frac{25\times325}{1100}\).
Calculate \(\frac{25\times325}{1100}=\frac{8125}{1100}\approx7.39\). So \(v_2\approx - 7.39\space m/s\). The negative sign indicates the direction is opposite to the direction of the turkey's motion (west).

Answer:

The recoil velocity of the launcher is approximately \(7.39\space m/s\) west.