Question

A bullet of mass $20\,g$ moving with a velocity of $300\dfrac{m}{s}$ gets embedded in a freely suspended wooden block of mass $880\,g$. What is the velocity acquired by the blocks?

Answer 1

Hint: The rate of change of an object's direction with respect to a frame of reference is its velocity, which is a function of time. A determination of an object's speed and direction of motion is referred to as velocity. Speed is the rate at which an object moves down a path in terms of distance, while velocity is the rate and direction of travel.

Complete step by step answer:
One of the most well-known laws of physics is the conservation of momentum. The conservation of momentum theorem states that a system's total momentum is still conserved.The law of momentum conservation states that unless an external force is applied, the total momentum of two or more bodies in an isolated system acting on each other remains constant. As a result, neither the creation nor the destruction of momentum is possible.

The theory of momentum conservation is a direct result of Newton's third law of motion.Newton's third law states that when an object A exerts a force on an object B, object B responds with a force of equal magnitude but opposite direction. Newton derived the law of conservation of momentum from this concept.

Bullet:
Convert the unit of mass to kilogram
${m_1} = 20\,g \\
\Rightarrow {m_1}= \dfrac{{20}}{{1000}}\,kg \\
\Rightarrow {m_1}= \dfrac{1}{{50}}\,kg$
Let the initial velocity be
\[{u_1} = 300\,m/s\]
Wooden Block: Let the mass of second object be
${m_2} = 880\,g = 880 \times {10^{ - 3}}\,kg$
Let its initial velocity be
\[{u_2} = 0\,m/s\]

To find the velocity of blocks with bullets embedded in it. Upon applying principle of conservation of linear momentum,
$\left( {{m_1} + {m_2}} \right)v = {m_1}{u_1} + {m_2}{u_2}{\text{ [Since }}{{\text{v}}_1} \\
\Rightarrow \left( {{m_1} + {m_2}} \right)v= {{\text{v}}_2}]$
$\Rightarrow (20 + 880) \times {10^{ - 3}} \times v = \dfrac{1}{{50}} \times 300 + 0$
$\Rightarrow (20 + 880) \times {10^{ - 3}} \times v = 6$
$\Rightarrow \dfrac{{900}}{{1000}}v = 6{\text{ or }}v = \dfrac{{6 \times 1000}}{{900}} \\
\therefore v= 6.67\;{\text{m}}/{\text{s}}$

Hence, the velocity acquired by the blocks are $6.67\;{\text{m}}/{\text{s}}$.

Note: Newton's third law states that if object A exerts a force on object B, object B must respond with a force of equal magnitude and in the opposite direction. This rule is often referred to as action-reaction, with the action being the force applied and the reaction being the force encountered as a result.