A man fires a bullet of mass $200\,gm$ at a speed of $5\,m/s$. The gun is of one $kg$ mass. By what velocity the gun rebounds backward?
- 1 m/s
- 0.01 m/s
- 0.1 m/s
- 10 m/s
The Correct Option is A
Solution and Explanation
Speed of bullet $(v_1) = 5$ m/sec. and mass of gun $(m_2) = 1 \,kg.$ Before firing, total momentum is zero.
$\therefore$ After firing total momentum is $m_1 v_1 + m_2 v_2$
From the law of conservation of momentum $m_1 v_1 + m_2 v_2=0$
or $v_2=\frac{-m_1v_1}{m_2}$
$=\frac{-0.2\times5}{1}$
$-1 \,m/sec.$
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Concepts Used:
Laws of Motion
The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.
Newton’s First Law of Motion
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion
Newton's 2nd law of motion deals with the relation between force and acceleration. According to the second law of motion, the acceleration of an object as built by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton’s Third Law of Motion
Newton's 3rd law of motion states when a body applies a force on another body that there is an equal and opposite reaction for every action.