Question

When a force of 1 N acts on 1 kg of mass at rest for 1 second, its final momentum is P. When 1 N force acts on 1 kg mass at rest through a distance of 1 m, its final momentum is P’. the ratio of P over P’ is
A. \[1:1\]
B. \[1:\sqrt{2}\]
C. \[1:2\]
D. \[2:1\]

Answer 1

Hint: In this question we have been asked to calculate the ratio of P over P’. Therefore, to solve this question, we shall first calculate the value of P and P’ from the given condition. We know that momentum is given as the product of velocity and mass of the object. In this case, the mass of the particle at both instances is the same therefore, the momentum will depend directly on the velocity of the particle at both instances.
Formula Used:
\[\Delta P=F\times t\]
\[F=ma\]
\[{{v}^{2}}-{{u}^{2}}=2as\]
\[P'=mv\]

Complete answer:
In this question we have to calculate the ratio of P over P’
Therefore, first solving for P
We know that,
\[\Delta P=F\times t\]
It is given that force applied is 1 N and t = 1s
Therefore,
\[{{P}_{f}}-{{P}_{i}}=1\]
The mass is initially at rest
Therefore,
\[{{P}_{f}}=P=1\] ………………. (A)
Now solving for P’,
We know that,
\[F=ma\]
Therefore, from given values
\[a=\dfrac{1N}{1kg}\]
Therefore,
\[a=1m/{{s}^{2}}\]
Now, from equation of motion
We know
\[{{v}^{2}}-{{u}^{2}}=2as\]
After substituting the values,
\[{{v}^{2}}=(2)(1)(1)\]
Therefore,
\[v=\sqrt{2}\]
We know that momentum is given by,
\[P'=mv\]
Substituting the values
\[P'=1\times \sqrt{2}\]
Therefore,
\[P'=\sqrt{2}\] ………………… (B)
Taking the ratio of (A) and (B)
We get,
\[\dfrac{P}{P'}=\dfrac{1}{\sqrt{2}}\]

Therefore, the correct answer is option B.

Note:
We know that momentum is the measure of how much mass is in how much motion. Therefore, momentum is given as the product of mass and velocity of the system. The unit of momentum in SI is kg m/s. every moving object with a certain mass has momentum. Momentum of a particle depends on the frame of reference. However, in any inertial frame the momentum of the particle is always conserved.