Question

A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is

• A. 2 cm/sec
• B. 20 cm/sec
• C. 200 cm/sec
• D. 2000 cm/sec

Answer 1

The correct option is C

200 cm/sec

Step1: Given data

A force of 100 dynes acts on mass of 5 gm for 10 sec

Step2: Formula used

$$\begin{gathered} F=\frac{dp}{dt} \\ p=mv \end{gathered}$$

Here, $F$ is the force, $p$ is the momentum, $t$ is the time,$m$ is the mass, and $v$ is the velocity.

Step3: Calculating velocity

We know that the force is equal to the rate of change of momentum, that is
$F=\frac{dp}{dt}\dots \dots \dots \dots \dots \dots \dots \dots ..\left(i\right)$
We also know that the momentum,$p$ is given by
$p=mv\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(ii\right)$
Substituting (ii) in (i)
$F=\frac{d\left(mv\right)}{dt}$

For a constant mass, we have
$F=m\frac{dv}{dt}$

Multiplying with $dt$ both the sides
$mdv=Fdt$
Dividing by $m$ on both sides
$dv=\frac{F}{m}dt$

Integrating both the sides

${\int }_{v_1}^{v_2}dv={\int }_{t_1}^{t_2}\frac{F}{m}dt$

As the force is constant for the whole time of motion, we have

$$\begin{gathered} {\int }_{v_1}^{v_2}dv=\frac{F}{m}{\int }_{t_1}^{t_2}dt \\ {\left[v\right]}_{v_1}^{v_2}=\frac{F}{m}{\left[t\right]}_{t_1}^{t_2} \end{gathered}$$

Substituting the limits, we get
$$\begin{gathered} v_2-v_1=\frac{F}{m}(t_2-t_1) \\ \Delta v=\frac{F}{m}\Delta t\dots \dots \dots \dots \dots \dots \dots \left(iii\right) \end{gathered}$$
So, this is the expression for the velocity produced.
Now, according to the question $F=100dynes$
Also, $\Delta t=10s,andm=5g$

So, from (iii), we get the velocity produced as $\Delta v=\frac{100}{5}\times 10$

On solving, we get $\Delta v=200cm/s$

Therefore the velocity produced is 200 cm/sec.

Hence, option C is the correct answer.