
For constant non zero mass, momentum versus velocity graph is a ?
A. straight line parallel to x-axis
B. straight line parallel to y-axis
C. straight line passing through origin
D. straight line passing through y-axis with an intercept
Answer
529.2k+ views
Hint:The point to be kept in mind while solving this question is that the momentum of a body is directly proportional to its velocity, keeping its mass to be constant. And the equation which is a part for this is,
\[{\text{y = mx}}\] ………..(1)
Which is also called the slope equation of tangent equation which we will use in our solution.
Complete step-by-step answer:
In Newtonian mechanics, momentum is the product of the mass and velocity of a body. If m is mass of an object and be its velocity, then the body’s momentum p is :-
${\text{p = mv}}$ ………..(2)
Moreover, Newton’s second law of motion states that the rate of change of a body’s momentum is equal to the net force acting on it. Momentum depends upon the frame of reference but in any inertial frame, it is a conserved quantity. If $\Delta {\text{p}}$ be the change in momentum and $\Delta {\text{t}}$ be the time interval, then the net force acting on the body can be given by :-
${\text{F = }}\dfrac{{\Delta {\text{p}}}}{{\Delta t}}$ ………..(3)
It is clear from equation (1) and (2), that momentum varies linearly with respect to velocity. And we know that the graph between two linearly varying quantities is a ‘straight line passing through the origin’. While the slope of the graph depends upon the mass of the body.
So, (C) option is correct.
Note:It is important to note that momentum is that, for a closed system, the total momentum is constant. This is known as the ‘ law of conservation of momentum’. This law applies to all interactions, including collisions, no matter how complicated the force is between the particles.
\[{\text{y = mx}}\] ………..(1)
Which is also called the slope equation of tangent equation which we will use in our solution.
Complete step-by-step answer:
In Newtonian mechanics, momentum is the product of the mass and velocity of a body. If m is mass of an object and be its velocity, then the body’s momentum p is :-
${\text{p = mv}}$ ………..(2)
Moreover, Newton’s second law of motion states that the rate of change of a body’s momentum is equal to the net force acting on it. Momentum depends upon the frame of reference but in any inertial frame, it is a conserved quantity. If $\Delta {\text{p}}$ be the change in momentum and $\Delta {\text{t}}$ be the time interval, then the net force acting on the body can be given by :-
${\text{F = }}\dfrac{{\Delta {\text{p}}}}{{\Delta t}}$ ………..(3)
It is clear from equation (1) and (2), that momentum varies linearly with respect to velocity. And we know that the graph between two linearly varying quantities is a ‘straight line passing through the origin’. While the slope of the graph depends upon the mass of the body.
So, (C) option is correct.
Note:It is important to note that momentum is that, for a closed system, the total momentum is constant. This is known as the ‘ law of conservation of momentum’. This law applies to all interactions, including collisions, no matter how complicated the force is between the particles.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

How do you find the composite numbers from 1 to 50 class 11 maths CBSE
