Question

A constant force (F) is applied on a
stationary particle of mass ‘m'. The
velocity attained by the particle in a
certain displacement will be proportiona
to
a) m
b) 1/m
c) 1/2^m
d)1/1/2^m

1/2^m = Square root m
​

Answer 1

Correct Question:-

A constant force $\displaystyle\sf {F}$ is applied on a stationary particle of mass $\displaystyle\sf {m.}$ The velocity attained by the particle in a certain displacement will be proportional to,

$\displaystyle\sf {(a)\quad\!m}$

$\displaystyle\sf {(b)\quad\!\dfrac {1}{m}}$

$\displaystyle\sf {(c)\quad\!\sqrt m}$

$\displaystyle\sf {(d)\quad\!\dfrac {1}{\sqrt m}}$

Answer:-

$\displaystyle\Large\boxed {\sf {\quad(d)\quad\!\dfrac {1}{\sqrt m}\quad}}$

Solution 1:-

By Second Law of Motion, we know that,

$\displaystyle\longrightarrow\sf{F=ma}$

where $\displaystyle\sf {a}$ is the acceleration of the body due to the force applied.

Then,

$\displaystyle\longrightarrow\sf{a=\dfrac {F}{m}\quad\quad\dots (1)}$

By third equation of motion,

$\displaystyle\longrightarrow\sf{v^2=u^2+2as}$

Since the particle was at rest,

$\displaystyle\longrightarrow\sf{v^2=2as\quad[\,u=0\,]}$

From (1),

$\displaystyle\longrightarrow\sf{v^2=2\cdot\dfrac {F}{m}\cdot s}$

At a certain displacement, since the force applied is constant,

$\displaystyle\longrightarrow\sf{v^2\propto\dfrac {1}{m}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {v\propto\dfrac {1}{\sqrt m}}}}$

Solution 2:-

Let the velocity depend upon mass of the particle, force applied and the displacement travelled as follows:

$\displaystyle\longrightarrow\sf{v\propto m^a\,F^b\,s^c\quad\quad\dots (2)}$

Taking dimensions of each term,

$\displaystyle\longrightarrow\sf{[v]=[m]^a\,[F]^b\,[s]^c}$

$\displaystyle\longrightarrow\sf{LT^{-1}=M^a\left(MLT^{-2}\right)^b\,L^c}$

$\displaystyle\longrightarrow\sf{M^0L^1T^{-1}=M^{a+b}\,L^{b+c}\,T^{-2b}}$

On equating corresponding dimensions, we get,

• $\displaystyle\sf {a=-\dfrac {1}{2}}$

• $\displaystyle\sf {b=\dfrac {1}{2}}$

• $\displaystyle\sf {c=\dfrac {1}{2}}$

Then (2) becomes,

$\displaystyle\longrightarrow\sf{v\propto\sqrt{\dfrac {Fs}{m}}}$

At a certain displacement, since the force applied is constant,

$\displaystyle\longrightarrow\sf{\underline {\underline {v\propto\dfrac {1}{\sqrt m}}}}$

Answer 2

Explanation:

$\huge\bold\red{ANSWER}$

A constant force (F) is applied on a stationary particle of mass m. The velocity attained by theparticle in a certain displacement will beproportional to

Acceleration, a= mF

F is constant here.

Using equation of motion as

v 2 −u 2 =2as

∵u=0

v 2 =2 F/ms

v∝ 1/√m

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