Conservative force: A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and not on the nature of path followed between the initial and final positions. Non-conservative forces: A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path between the initial and final positions. Examples of Conservative Forces: - Gravitational force, not only due to the Earth but in its general form as given by the universal law of gravitation, is a conservative force.
- Elastic force in a stretched or compressed spring is a conservative force.
- Electrostatic force between two stationary electric charges is a conservative force.
- Magnetic force between two magnetic poles is a conservative force.
Properties of Conservative Forces: - Work done by or against a conservative force depends only on the initial and final positions of the body.
- Work done by or against a conservative force does not depend upon the nature of the path between initial and final positions of the body.
- Work done by or against a conservative force in a round trip is zero.
- The work done by a conservative force is completely recoverable.
Numerical: Mass of rifle, M = 5 kg; Mass of bullet, m = 5g = 5 × 10-3 kg Speed of bullet, v = 500 ms-1 - Kinetic energy acquired by bullet,
$\text{K}_1=\frac{1}{2}\text{m}\upsilon^2=\frac{1}{2}\times5\times10^{-3}\times(500)^2=625\text{ J}$
- According to the law of conservation of momentaum (As no external force acts on the bullet or rifle)
$0=\text{MV}+\text{m}\upsilon$
$\therefore$ V (speed of rifle) $=\frac{\text{m}\upsilon}{\text{M}}=-\frac{5\times10^{-3}\times500}{5}$
⇒ V = -0.5 ms-1
$-\upsilon\text{e}$ sign shows that rifle recoils back.
$\therefore$ Kinetic energy of rifle $\frac{1}{2}\text{M}\text{V}^2$
$\frac{1}{2}\times5\times(-0.5)^2=0.625\text{ J}$
- Since $\text{m}\upsilon+\text{MV}= 0$
or $\text{m}\frac{\text{dx}_1}{\text{dt}}+\text{M}\frac{\text{dx}_2}{\text{dt}}=0$
$ \begin{bmatrix}\text{Q}\text{V}=\frac{\text{dx}_1}{\text{dt}}=\frac{\text{Distance moved by bullet}}{\text{times}}\\\text{V}=\frac{\text{dx}_2}{\text{dt}}=\frac{\text{Distance moved by rifle}}{\text{time}}\end{bmatrix}$
$\therefore$ mdx1 + Mdx2 = 0
$\Big|\frac{\text{dx}_2}{\text{dx}_1}\Big|=\frac{\text{m}}{\text{M}}=\frac{5\times10^{-3}}{5}=\frac{1}{1000}$.
