
How do you calculate the change in momentum of an object?
Answer
546.9k+ views
Hint: Momentum is the vector quantity which can be described by Newton’s second law of motion and can be found by the product of the velocity and mass of an object.
Formula used:
$\begin{align}
& \Delta P=m.\text{ }\Delta v \\
& \Delta P=F.\Delta t \\
\end{align}$
Complete step by step solution:
There are two ways two find the change in the momentum.
1) The change in the momentum is equal to the product of the mass and change in the velocity.
$\begin{align}
& \Rightarrow \Delta P=m.\Delta x \\
& \Rightarrow \Delta P=mx\left( {{v}_{f}}-{{v}_{i}} \right)....\left( 1 \right) \\
\end{align}$
Where,
$\Delta P=$ change in the momentum.
m = mass of an object
∆v = change in the velocity
${{v}_{f}}$ = final velocity of an object.
${{v}_{i}}$= initial velocity of an object.
2) In this method we can find change in momentum by using Newton’s second law of motion.
Newton’s second law of motion:-
The time rate of change of momentum is equal to the force acting on the particle.
If the force is constant in the given time interval then the product of force and change in time will give us the change in the momentum.
$\Delta P=F.\Delta t...\left( 2 \right)$
Where,
$\Delta P$= change in the momentum
F = force acting on the object.
$\Delta t$= time intervals or change in time.
Note:
In the method 2, be careful while putting the signs of final and the initial velocity of an object. i.e. A 2kg mass initially moving with 5m/s to the right and rebounds off the wall and starts travelling in left direction with the final velocity of 3m/s.
Take right as +ve, and left as –ve,
Then, ${{v}_{f}}$= -3cm/s
${{v}_{i}}$ = +5m/s
m = 2kg.
From the equation (1)
$\begin{align}
& \Rightarrow \Delta P=2\left( -3-5 \right) \\
& =-16kgm/s \\
\end{align}$
This will clarify about signs of the velocity.
Formula used:
$\begin{align}
& \Delta P=m.\text{ }\Delta v \\
& \Delta P=F.\Delta t \\
\end{align}$
Complete step by step solution:
There are two ways two find the change in the momentum.
1) The change in the momentum is equal to the product of the mass and change in the velocity.
$\begin{align}
& \Rightarrow \Delta P=m.\Delta x \\
& \Rightarrow \Delta P=mx\left( {{v}_{f}}-{{v}_{i}} \right)....\left( 1 \right) \\
\end{align}$
Where,
$\Delta P=$ change in the momentum.
m = mass of an object
∆v = change in the velocity
${{v}_{f}}$ = final velocity of an object.
${{v}_{i}}$= initial velocity of an object.
2) In this method we can find change in momentum by using Newton’s second law of motion.
Newton’s second law of motion:-
The time rate of change of momentum is equal to the force acting on the particle.
If the force is constant in the given time interval then the product of force and change in time will give us the change in the momentum.
$\Delta P=F.\Delta t...\left( 2 \right)$
Where,
$\Delta P$= change in the momentum
F = force acting on the object.
$\Delta t$= time intervals or change in time.
Note:
In the method 2, be careful while putting the signs of final and the initial velocity of an object. i.e. A 2kg mass initially moving with 5m/s to the right and rebounds off the wall and starts travelling in left direction with the final velocity of 3m/s.
Take right as +ve, and left as –ve,
Then, ${{v}_{f}}$= -3cm/s
${{v}_{i}}$ = +5m/s
m = 2kg.
From the equation (1)
$\begin{align}
& \Rightarrow \Delta P=2\left( -3-5 \right) \\
& =-16kgm/s \\
\end{align}$
This will clarify about signs of the velocity.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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

