Question

1. a 0.060 kg rifle bullet leaves the muzzle with a velocity of 6.0 x 10² m/s. the 3.0 kg rifle is held firmly by a 60.0 kg man. with what initial velocity will the man and rifle recoil? compare your answer with the answer to the sample problem preceding these exercises.

Explanation:

Step1: Apply law of conservation of momentum

The initial momentum of the system (man + rifle + bullet) is zero. Let the mass of the bullet be $m_b = 0.060$ kg, its velocity be $v_b=6.0\times 10^{2}$ m/s, the mass of the rifle be $m_r = 3.0$ kg and the mass of the man be $m_m=60.0$ kg. Let the recoil velocity of the man - rifle system be $v$. According to the law of conservation of momentum $0 = m_bv_b+(m_m + m_r)v$.

Step2: Solve for the recoil velocity $v$

We can re - arrange the equation $0 = m_bv_b+(m_m + m_r)v$ to solve for $v$. First, we get $(m_m + m_r)v=-m_bv_b$. Then $v =-\frac{m_bv_b}{m_m + m_r}$. Substitute $m_b = 0.060$ kg, $v_b = 6.0\times 10^{2}$ m/s, $m_m = 60.0$ kg and $m_r = 3.0$ kg into the formula. $v=-\frac{0.060\times6.0\times 10^{2}}{60.0 + 3.0}$. Calculate the numerator: $0.060\times6.0\times 10^{2}=36$. Calculate the denominator: $60.0+3.0 = 63.0$. So $v=-\frac{36}{63.0}\approx - 0.57$ m/s. The negative sign indicates that the direction of the recoil is opposite to the direction of the bullet's motion.

Answer:

The initial velocity of the man and rifle recoil is approximately $0.57$ m/s in the direction opposite to the bullet's motion.