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Question

Question: A force of \(100\)dynes acts on mass of \(5\)gm for\(10\) seconds. The velocity produced is (A) \(...

A force of 100100dynes acts on mass of 55gm for1010 seconds. The velocity produced is
(A) 2cm/s2cm/s
(B) 20cm/s20cm/s
(C) 200cm/s200cm/s
(D) 2000cm/s2000cm/s

Explanation

Solution

Hint To solve this question, we have to use the basic definition of force. Using the given values in the expression of force thus obtained, we will get the required answer.
Formula Used The formula used to solve this question is
F=dpdtF = \dfrac{{dp}}{{dt}}
p=mvp = mv
Here,FF is the force, pp is the momentum, tt is the time, mm is the mass, and vv is the velocity.

Complete step-by-step solution
We know that the force is equal to the rate of change of momentum, that is
F=dpdtF = \dfrac{{dp}}{{dt}} ……………………..(i)
We also know that the momentum, pp is given by
p=mvp = mv ………………………...(ii)
Substituting (ii) in (i)
F=d(mv)dtF = \dfrac{{d\left( {mv} \right)}}{{dt}}
For a constant mass, we have
F=mdvdtF = m\dfrac{{dv}}{{dt}}
Multiplying with dtdt both the sides
mdv=Fdtmdv = Fdt
Dividing by m on both sides
dv=Fmdtdv = \dfrac{F}{m}dt
Integrating both the sides
v1v2dv=t1t2Fmdt\int\limits_{v1}^{v2} {dv} = \int\limits_{t1}^{t2} {\dfrac{F}{m}dt}
As the force is constant for the whole time of motion, we have
v1v2dv=Fmt1t2dt\int\limits_{v1}^{v2} {dv} = \dfrac{F}{m}\int\limits_{t1}^{t2} {dt}
[v]v1v2=Fm[t]t1t2\left[ v \right]_{v1}^{v2} = \dfrac{F}{m}\left[ t \right]_{t1}^{t2}
Substituting the limits, we get
v2v1=Fm(t2t1)v2 - v1 = \dfrac{F}{m}\left( {t2 - t1} \right)
Or, Δv=FmΔt\Delta v = \dfrac{F}{m}\Delta t ………………...(iii)
So, this is the expression for the velocity produced.
Now, according to the question
F=100 dynesF = 100{\text{ }}dynes
Also,
Δt=10s\Delta t = 10s , and
m=5gm = 5g
So, from (iii), we get the velocity produced as
Δv=1005×10\Delta v = \dfrac{{100}}{5} \times 10
On solving, we get
Δv=200cm/s\Delta v = 200cm/s

Hence, the velocity produced is equal to 200cm/s200cm/s

Note On looking at the units of the quantities given in the question, we may attempt to convert them all into the respective SI units. But we should take care of the fact that all of the fact that all the units given belong to the C.G.S. system of units. Also, the units of velocity given in the options belong to the C.G.S. system. So, we do not need to perform any conversion at all. Making conversions will only increase our load. Hence, before making any conversion of units, we must first check the system of units the respective units belong to.