Question

A body of mass \[2kg\] moves an acceleration \[3m/{s^2}\]. The change in momentum in \[1sec\] is
A) \[\dfrac{2}{3}kgm/s\]
B) \[\dfrac{3}{2}kgm/s\]
C) \[6kgm/s\]
D) None of these

Answer 1

Hint:The above problem is related to the law of motion in which we read three laws of motion that satisfy the motion of any moving particle.
When a body moves with some acceleration then a force starts working on it. So, we can easily apply Newton's second law of motion. In other words ‘the force applied to a body is equal to the product of mass of the body and acceleration produced in the body.’

Complete step by step solution: -
When a body moves with an acceleration then according to the second law of motion. A force starts acting on it, so if the mass of the body is \[2kg\] and acceleration is \[3m/{s^2}\]. So, we can find force by applying $F = ma$, we have
$F = 2 \times 3$
$ \Rightarrow F = 6N$ ……………..(i)
But actually the force applied to the body is proportional to the change in momentum. i.e.
$F = \dfrac{{\Delta p}}{{\Delta t}}$ ………………..(ii)
In the question, we are finding change in momentum in one second.
So,
$
  \Delta t = 1\sec \\
  F = 6N \\
  \Delta p = ? \\
$
Putting the values in equation (ii)
$
   \Rightarrow 6 = \dfrac{{\Delta p}}{1} \\
   \Rightarrow \Delta p = 6 \times 1 \\
   \Rightarrow \Delta p = 6kgm/s \\
$
Hence change in momentum in one second is \[6kgm/s\].

Therefore, option C is correct.

Additional information:
According to the second law of motion given by Newton, ‘The rate of change of momentum is directly proportional to the applied force.’ i.e.
$F \propto \dfrac{{\Delta p}}{{\Delta t}}$ ……………..(a)
When we remove proportionality, a constant is used instead of that.
$F = k\dfrac{{\Delta p}}{{\Delta t}}$
But momentum of any moving body with velocity $v$ is $p = mv$
Putting the value of $p$ in above equation, we get-
$F = k\dfrac{{\Delta \left( {mv} \right)}}{{\Delta t}}$
The mass of a body is a constant quantity. So,
$F = km\dfrac{{\Delta v}}{{\Delta t}}$
Where $\dfrac{{\Delta v}}{{\Delta t}}$ represents the acceleration of the body as acceleration is equal to the rate of change in velocity with respective time and represented by $a$
So, $F = kma$
By experiments $k = 1$
So, $F = ma$ ………………….(b)
Hence, we can write newton’s second law as ‘the force applied to a body is equal to the product of mass of the body and acceleration produced in the body.’

Note: - This question follows Newton's second law of motion. It should be remembered that acceleration means the velocity is increasing or decreasing. If increasing then change in velocity is considered as acceleration and taken as positive magnitude. But if the velocity decreases, then change in velocity is considered retardation and taken as negative magnitude.