Question

A Diwali rocket is ejecting 0.05 kg of gases per second at velocity of \[400\,m{s^{ - 1}}\]. The accelerating force on the rocket is
A. 20 dyne
B. 20 newton
C. 20 kg wt
D. Sufficient data not given

Answer 1

Hint: Use Newton’s second law in terms of rate of change of linear momentum. Check whether the units are given in S.I. units or CGS units.

Formula used:
\[F = \dfrac{{dP}}{{dt}}\]
Here, P is the linear momentum.

Complete step by step answer:
We have, according to Newton’s second law of motion, the force acting on the body of mass m moving with velocity v is equal to the rate of change of linear momentum of the body.
\[F = \dfrac{{dP}}{{dt}}\]
Here, P is the linear momentum of the particle.
The linear momentum of the rocket moving with velocity v is,
\[P = mv\]
Therefore, the force on the Diwali rocket will be,
\[F = \dfrac{{d\left( {mv} \right)}}{{dt}}\]
Since the velocity of Diwali rocket is constant, we can write the above equation as follows,
\[F = v\dfrac{{dm}}{{dt}}\]
We have given, the rate of ejection of mass of gases is 0.05 kg per second. Therefore, substitute 0.05 kg per sec for m and \[400\,m{s^{ - 1}}\] in the above equation.
\[F = \left( {400\,m{s^{ - 1}}} \right)\left( {0.05\,kg\,{s^{ - 1}}} \right)\]
\[ \Rightarrow F = 20\,kg\,m{s^{ - 2}}\]
\[ \Rightarrow F = \left( {20\,kg\,m{s^{ - 2}}} \right)\left( {\dfrac{{1\,N}}{{1\,kg\,m{s^{ - 2}}}}} \right)\]
\[ \Rightarrow F = 20\,N\]
Dyne is the CGS unit of force, but the quantities given in the question are in S.I. units. Therefore, the option (A) is incorrect.

So, the correct answer is “Option B”.

Note:
We can also use the equation, \[F = ma\], where a is the acceleration. The acceleration of the body is given as, \[a = \dfrac{{{v_2} - {v_1}}}{{\Delta t}}\]. On substituting the expression for acceleration in the expression for force, we will find the expression for rate of change in linear momentum.