Question 118: Laws of Motion

Question

A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = $$-$$ $$\alpha$$x2. The distance at which the particle stops :

Accepted Answer

Given, speed of projection = v
0
Damping force, F = ma = $$-$$ $$\alpha$$x
2
$$\Rightarrow$$ a = $$-$$ $$\alpha$$x
2
/ m
Also, $$a = v{{dv} \over {dx}}$$
$$ \Rightarrow vdv = a\,dx =  - {\alpha  \over m}{x^2}dx$$
Integrating both sides, we get
$$\int_{{v_0}}^v {vdv = \int_0^x { - {\alpha  \over m}{x^2}dx} } $$
$$ \Rightarrow \left( {{{{v^2}} \over 2}} \right)_{{v_0}}^0 =  - {\alpha  \over m}\left( {{{{x^3}} \over 3}} \right)_0^x$$
$$ \Rightarrow 0 - v_0^2/2 =  - {\alpha  \over m}{{{x^3}} \over 3} \Rightarrow x = {\left( {{{3m} \over 2}{{v_0^2} \over \alpha }} \right)^{1/3}}$$

Answer

$${\left[ {{{3mv_0^2} \over {2\alpha }}} \right]^{{1 \over 3}}}$$

Answer

$${\left( {{{2{v_0}} \over {3\alpha }}} \right)^{{1 \over 3}}}$$

Answer

$${\left( {{{3v_0^2} \over {2\alpha }}} \right)^{{1 \over 2}}}$$

Answer

$${\left( {{{2v_0^2} \over {3\alpha }}} \right)^{{1 \over 2}}}$$

Laws of Motion

A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = $- \alpha$x2. The distance at which the particle stops :

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