Question

A man standing on very slick ice fires a rifle horizontally. The mass of the man together with the rifle is 70kg , and the mass of the bullet is 10g . If the bullet leaves the muzzle at a speed of 500m/s , what is the final speed of the man?

Answer 1

Concepts and reason

The concept required to solve the given problem is the law of conservation of linear momentum.

Calculate the velocity of man with the help of law of conservation of linear momentum.

Fundamentals

Law of conservation of linear momentum: It states that when no external force acts on a body the momentum remains conserved. Mathematically the statement may be expressed as,

${p_i} = {p_f}$

Here, ${p_i}{\rm{ and }}{p_f}$ is the initial and final momentum of the body into consideration.

Momentum is given by,

$p = mv$

Here, $m$ is the mass and $v$ is the velocity of the body.

The initial velocity of the man with rifle is,

${u_1} = 0{\rm{ m /s }}$

The initial velocity of the bullet is,

${u_2} = 0{\rm{ m / s}}$

Apply the law of conservation of linear momentum.

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$ is the mass of the man and rifle together,${m_2}$ is the mass of bullet, ${u_1}$ is the initial velocity of the man and the rifle together, ${u_2}$ is the initial velocity of the bullet, ${v_1}$ is the final velocity of the man and rifle together and ${v_2}$ is the final velocity of the bullet.

Rearrange the above equation.

${v_1} = \frac{{{m_1}{u_1} + {m_2}{u_2} - {m_2}{v_2}}}{{{m_1}}}$ ……(1)

Substitute $0{\rm{ m / s}}$ for ${u_1}{\rm{ and }}{u_2}$, $70{\rm{ kg}}$ for ${m_1}$, $10{\rm{ g}}$ for ${m_2}$, and $500{\rm{ m / s}}$ for ${v_2}$ in equation (1).

$\begin{array}{c}\\{v_1} = \frac{{{m_1}{u_1} + {m_2}{u_2} - {m_2}{v_2}}}{{{m_1}}}\\\\ = \frac{{\left( {70{\rm{ kg}}} \right)\left( {0{\rm{ m / s}}} \right) + \left( {10{\rm{ g}}} \right)\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right)\left( {0{\rm{ m / s}}} \right) - \left( {10{\rm{ g}}} \right)\left( {\frac{{1{\rm{ kg}}}}{{1000g}}} \right)\left( {500{\rm{ m / s}}} \right)}}{{\left( {70{\rm{ kg}}} \right)}}\\\\ = 0.0714{\rm{ m / s}}\\\end{array}$

Ans:

The final speed of the man is $0.0714{\rm{ m / s}}$.