
Give the relation between acceleration, mass and force.
Answer
489.9k+ views
Hint: The relation between the above three quantities is given by Newton’s second law. First we will state Newton's second law. Further we will express the law to its mathematical form and obtain the relation between acceleration, mass and force.
Formula used:
$\dfrac{dp}{dt}=\dfrac{d(mv)}{dt}$
Complete step-by-step solution:
To begin with let us first define Newton’s second law of motion.
Newton’s second law of motion states that the rate of change of momentum of a body is equal to the force acting on the body in a particular direction. If we consider the body to move along a straight line such that its momentum increases then we can say that the direction of force acting is along the direction of the momentum vector. Similarly if the force acting is in the opposite direction to that of the momentum vector of the body then its momentum decreases.
Now let us understand the above statement mathematically. For that let us consider a body of mass ‘m’ and moves with velocity ‘v’. Therefore the momentum ‘p’ of the body is given by,
$p=mv$
From Newton’s law of motion we now know that the rate of change of momentum of a body is equal to the force acting on the body. Hence the force F is equal to,
$\begin{align}
& p=mv \\
& F=\dfrac{dp}{dt}=\dfrac{d(mv)}{dt} \\
& \Rightarrow F=v\dfrac{dm}{dt}+m\dfrac{dv}{dt}.....(1) \\
\end{align}$
The acceleration ‘a’ of a body is defined as the rate of change of velocity i.e. $\dfrac{dv}{dt}$. And since mass does not change with time equation 1 can be written as,
$\begin{align}
& F=v\dfrac{dm}{dt}+m\dfrac{dv}{dt} \\
& \Rightarrow F=v(0)+ma \\
& \therefore F=ma \\
\end{align}$
Hence the required relation between acceleration, mass and force is given by $F=ma$.
Note:For a non relativistic particle the speed of the particle is very small compared to the speed of light. Hence the mass of the body remains constant. Even if in a particular scenario if the mass of the body changes as well then still our result holds true. This can be verified by looking at the dimensions of the quantities on both the sides of the equation.
Formula used:
$\dfrac{dp}{dt}=\dfrac{d(mv)}{dt}$
Complete step-by-step solution:
To begin with let us first define Newton’s second law of motion.
Newton’s second law of motion states that the rate of change of momentum of a body is equal to the force acting on the body in a particular direction. If we consider the body to move along a straight line such that its momentum increases then we can say that the direction of force acting is along the direction of the momentum vector. Similarly if the force acting is in the opposite direction to that of the momentum vector of the body then its momentum decreases.
Now let us understand the above statement mathematically. For that let us consider a body of mass ‘m’ and moves with velocity ‘v’. Therefore the momentum ‘p’ of the body is given by,
$p=mv$
From Newton’s law of motion we now know that the rate of change of momentum of a body is equal to the force acting on the body. Hence the force F is equal to,
$\begin{align}
& p=mv \\
& F=\dfrac{dp}{dt}=\dfrac{d(mv)}{dt} \\
& \Rightarrow F=v\dfrac{dm}{dt}+m\dfrac{dv}{dt}.....(1) \\
\end{align}$
The acceleration ‘a’ of a body is defined as the rate of change of velocity i.e. $\dfrac{dv}{dt}$. And since mass does not change with time equation 1 can be written as,
$\begin{align}
& F=v\dfrac{dm}{dt}+m\dfrac{dv}{dt} \\
& \Rightarrow F=v(0)+ma \\
& \therefore F=ma \\
\end{align}$
Hence the required relation between acceleration, mass and force is given by $F=ma$.
Note:For a non relativistic particle the speed of the particle is very small compared to the speed of light. Hence the mass of the body remains constant. Even if in a particular scenario if the mass of the body changes as well then still our result holds true. This can be verified by looking at the dimensions of the quantities on both the sides of the equation.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

