Informative line

Impulse

Learn impulse definition with problems and examples in physics, Practice average force impulse and force time graph, impulse momentum theorem.

Idea of Impulsive Situation

  • According to Newton's Second Law

\(\overrightarrow{F}_{net}=\dfrac{d\overrightarrow{P}}{dt}\)

\(\displaystyle\int\overrightarrow{F}_{net}.dt=d\overrightarrow{P}\)

\(\displaystyle\int\overrightarrow{F}_{net}.dt=\overrightarrow{\Delta P}\)

  • We call \(\displaystyle\int\overrightarrow{F}_{net}.dt\) as Impulse.

\(\displaystyle\int\overrightarrow{F}_{net}.dt=\overrightarrow{P}_f- \overrightarrow{P}_i\)

  • The above equation is another form of Newton's Second Law.
  • Sometimes, a very large force (generally variable) acts for a very short time.

            For e.g., a very hard ball collides with a wall.

  • In this situation, we can't calculate the force acting on the ball using Newton's Second Law.

  • Change in the shape of the ball is for such a very short interval of time that we can't see it with naked eyes.
  • During this collision, a large force (variable) acts on the ball.
  • We can't calculate this force, but velocity before and after collision can easily be calculated and hence the effect of force is 

 \(\displaystyle\int\overrightarrow{F}_{impulsive}.dt=m\overrightarrow{v'}- m\overrightarrow{v}\)

  • For such impulsive situation we can't calculate \(\overrightarrow{F}_{impulsive}\)
  • But we can calculate

                                  \(\displaystyle\int F_{imp}.dt\)

This quantity is known as Impulse.

Graphical Representation of Impulse

  \(\displaystyle\int F.dt\) = area under the curve

           \(\displaystyle\int F.dt=\Delta P\)

  • So for this impulsive situation, we can calculate the area under curve, hence the change in momentum.

Note : (1) When an impulsive force acts for a short time interval, neglect the effect of finite force for that time interval.

           (2) There are situations where forces are not impulsive, but still we can calculate impulse i.e., change in momentum.

Illustration Questions

Which one of the following is impulsive in nature?

A Shaking of hands

B Walking

C Hitting a punching bag

D Chin-up exercise

×

While shaking of hands no large force acts for a short duration. So, this is not impulsive in nature.

While walking no large force acts for a short duration. So, this is not impulsive in nature.

Hitting a punching bag exerts a large force for a very short duration. So, this is impulsive in nature.

While chin-up exercise no large force acts for a short duration. So, this is not impulsive in nature.

Which one of the following is impulsive in nature?

A

Shaking of hands

.

B

Walking

C

Hitting a punching bag

D

Chin-up exercise

Option C is Correct

Calculation of Velocity after Event using Impulse Momentum Theorem(Initial velocity is zero)

  • In some situations, value of impulsive force acting on a particle, cannot be calculated, but its effect can be measured by calculating \(\displaystyle\int\overrightarrow{F}.dt\). This quantity is termed as IMPULSE.
  • So, mathematically

                  Impulse \((\overrightarrow{J})=\displaystyle\int\overrightarrow{F}.dt\)  ... (1)

                  From Newton's Second Law

                      \(\Delta\overrightarrow{P}=\displaystyle\int\overrightarrow{F}.dt\)             ... (2)

                    From (1) and (2)

                     \(\overrightarrow{J}=\Delta\overrightarrow{P}\)

  • The change in the momentum of a particle is equal to the impulse of the net force acting on the particle.
  • \(\overrightarrow{P}_f - \overrightarrow{P}_i=\overrightarrow{J}\)    ... (3)

  • This is an important result and is known as the Impulse- Momentum Theorem.

Illustration Questions

A ball of mass 2 kg is at rest on a smooth horizontal surface. An impulse of \(10 \hat i\,N.s\)  is imparted to it. Find the final velocity of ball.

A \(10\,\hat i\;m/s\)

B \(6\,\hat i\;m/s\)

C \(5 \,\hat i\;m/s\)

D \(18\,\hat i\;m/s\)

×

Draw diagrams before and after the event.

image

Initial momentum = 0 = \(\overrightarrow{P}_i\)

Final momentum =   \(2 × \overrightarrow{v}\) = \(\overrightarrow{P}_f\)

Change in momentum

\(\Delta\overrightarrow{P}=\overrightarrow{P}_f - \overrightarrow{P}_i\)

        \(=2\overrightarrow{v}\)

\(\overrightarrow{J} = \Delta\overrightarrow{P}\)  (By Impulse Momentum Theorem)

\(10\,\hat i=2\overrightarrow{v}\)

\(\overrightarrow{v}=5\,\hat i\;m/s\)

A ball of mass 2 kg is at rest on a smooth horizontal surface. An impulse of \(10 \hat i\,N.s\)  is imparted to it. Find the final velocity of ball.

A

\(10\,\hat i\;m/s\)

.

B

\(6\,\hat i\;m/s\)

C

\(5 \,\hat i\;m/s\)

D

\(18\,\hat i\;m/s\)

Option C is Correct

Calculation of Velocity after Event using Impulse Momentum Theorem(Initial velocity is non zero)

  • In some situations, value of impulsive force acting on a particle, cannot be calculated, but its effect can be measured by calculating \(\displaystyle\int\overrightarrow{F}.dt\). This quantity is termed as IMPULSE.
  • So, mathematically

                  Impulse \((\overrightarrow{J})=\displaystyle\int\overrightarrow{F}.dt\)  ... (1)

                  From Newton's Second Law

                      \(\Delta\overrightarrow{P}=\displaystyle\int\overrightarrow{F}.dt\)             ... (2)

                    From (1) and (2)

                     \(\overrightarrow{J}=\Delta\overrightarrow{P}\)

  • The change in the momentum of a particle is equal to the impulse of the net force acting on the particle.
  • \(\overrightarrow{P}_f - \overrightarrow{P}_i=\overrightarrow{J}\)    ... (3)

  • This is an important result and is known as the Impulse- Momentum Theorem.

Illustration Questions

A ball of mass 2 kg is moving on a smooth horizontal surface with a velocity of \(10 \,\hat i\;m/s\). An impulse of  \(40\,\hat i\;N.s\) is imparted to it. Find the final velocity of ball.

A \(30\,\hat i\;m/s\)

B \(20\,\hat i\;m/s\)

C \(10\,\hat i\;m/s\)

D \(15 \,\hat i\;m/s\)

×

Draw diagrams of before and after the event.

image

Initial momentum =  \((2×10)\, \hat i =\overrightarrow{P}_i\)

\(\overrightarrow{P}_i = 20 \,\hat i \;kg\,m/s\)

Final momentum = \((2 ×\vec v)\,kg\,m/s\)

Change in momentum

\(\Delta\overrightarrow{P}=\overrightarrow{P}_f - \overrightarrow{P}_i\)

 

By Impulse-Momentum Theorem

\(\overrightarrow{J} = \Delta\overrightarrow{P}\)

\(40 \,\hat i=2\overrightarrow{v} - 20 \,\hat i\)

\(2\overrightarrow{v} = 60 \,\hat i\)

\(\overrightarrow{v}=30\,\hat i\,m/s\)

A ball of mass 2 kg is moving on a smooth horizontal surface with a velocity of \(10 \,\hat i\;m/s\). An impulse of  \(40\,\hat i\;N.s\) is imparted to it. Find the final velocity of ball.

A

\(30\,\hat i\;m/s\)

.

B

\(20\,\hat i\;m/s\)

C

\(10\,\hat i\;m/s\)

D

\(15 \,\hat i\;m/s\)

Option A is Correct

Qualitative Idea of Average Force \((\overrightarrow{F}_{avg})\)

Average force can be calculated for the time interval of an impulsive force as discussed below :

We know that

\(\overrightarrow{J}=\displaystyle\int\overrightarrow{F}.dt=\Delta \overrightarrow{P}\)

  • From Newton's Second Law

\(\overrightarrow{F}_{avg}=\dfrac{\Delta\overrightarrow{P}}{\Delta t}\) ... (1)

  • It can be concluded that for constant value of \(\Delta\overrightarrow{P}\), if \(\Delta t\) increases, \(\overrightarrow{F}_{avg}\) decreases and if \(\Delta t\) decreases \(\overrightarrow{F}_{avg}\) increases.
  • When we jump from a height on a hard surface (like floor), we get much severe injury as compared to the jump from same height on sand.
  • In both the cases, the change in momentum is same but duration of impact is more in the case of sand.
  • Thus, \(F_{avg}\) is less and we get less or no injury.

Illustration Questions

If while catching the ball, the time interval required to stop it, becomes one-third. Then the average force applied by the ball on hand will be

A Doubled

B Tripled

C Halved

D One-Third

×

As we know 

\(\overrightarrow{F}_{avg}=\dfrac{\Delta\overrightarrow{P}}{\Delta t}\) ... (1)

Now from equation (1), it can be seen that for equal change in momentum, \(\overrightarrow{F}_{avg}\propto\dfrac{1}{\Delta t}\).

So, if the time interval becomes one-third, the average force applied by ball on hand will be tripled.

If while catching the ball, the time interval required to stop it, becomes one-third. Then the average force applied by the ball on hand will be

A

Doubled

.

B

Tripled

C

Halved

D

One-Third

Option B is Correct

Calculation of Average Force (Favg)

Average force can be calculated for the time interval of an impulsive force as discussed below :

We know that

\(\overrightarrow{J}=\displaystyle\int\overrightarrow{F}.dt=\Delta \overrightarrow{P}\)

  • From Newton's Second Law

\(\overrightarrow{F}_{avg}=\dfrac{\Delta\overrightarrow{P}}{\Delta t}\) ... (1)

  • It can be concluded that for constant value of \(\Delta\overrightarrow{P}\), if \(\Delta t\) increases, \(\overrightarrow{F}_{avg}\) decreases and if \(\Delta t\) decreases \(\overrightarrow{F}_{avg}\) increases.
  • When we jump from a height on a hard surface (like floor), we get much severe injury as compared to the jump from same height on sand.
  • In both the cases, the change in momentum is same but duration of impact is more in the case of sand.
  • Thus, \(F_{avg}\) is less and we get less or no injury.

Calculation of Average Force \((\overrightarrow{F}_{avg})\)

Step 1 : Write down final momentum and initial momentum vectors.

Step 2 : Calculate change in momentum

                                 \(\Delta\overrightarrow{P}=\overrightarrow{P}_f - \overrightarrow{P}_i\)

Step 3 : \(\overrightarrow{F}_{avg}=\dfrac{\Delta\overrightarrow{P}}{\Delta t}\)   

Illustration Questions

A ball of mass 2 kg, moving with a velocity of \(10 \,\hat i\,m/s\) collides with another ball. If after collision velocity of the ball becomes \(6\, \hat i\,m/s\) and time duration of collision is 0.02 sec, then find average force on the ball during collision.

A \(600\, \hat i\,N\)

B \(-400 \,\hat i\,N\)

C \(100\, \hat i\,N\)

D \(-200\, \hat i\,N\)

×

Initial momentum =  \((2×10) \hat i\,kg\,m/s =\overrightarrow{P}_i\)

Final momentum = \((2×6) \hat i \,kg\,m/s=\overrightarrow{P}_f\)

Change in momentum

\(\Delta\overrightarrow{P}=\overrightarrow{P}_f - \overrightarrow{P}_i\)

        \(=(12-20)\hat i\,kg\,m/s\)

        \(=-8\,\hat i\,kg\,m/s\)             

Average force is given as

\(\overrightarrow{F}_{avg}=\dfrac{\Delta\overrightarrow{P}}{\Delta t}\)

\(=\dfrac{-8\,\hat i\,kg\,m/s}{0.02\,s}\)

\(=-400 \,\hat i\,N\)

A ball of mass 2 kg, moving with a velocity of \(10 \,\hat i\,m/s\) collides with another ball. If after collision velocity of the ball becomes \(6\, \hat i\,m/s\) and time duration of collision is 0.02 sec, then find average force on the ball during collision.

A

\(600\, \hat i\,N\)

.

B

\(-400 \,\hat i\,N\)

C

\(100\, \hat i\,N\)

D

\(-200\, \hat i\,N\)

Option B is Correct

Calculation of Impulse from Force-Time Graph

  • We know that

\(\overrightarrow{J}=\displaystyle\int\limits_{t_i}^{t_f}\overrightarrow{F}.dt\)

  • It can be seen that area under F - t curve gives the magnitude of impulse.

Illustration Questions

The force versus time graph of a particle, moving along the x-axis is given. Calculate the impulse of the force during time t = 2 s to t = 2.1 s.

A 20 N.s

B 25 N.s

C 30 N.s

D 40 N.s

×

Impulse is given as

\(|\overrightarrow{J}|=\displaystyle\int\limits_{t_1}^{t_2}F_x.dt\)

= Area under F - t curve

\(=\dfrac{1}{2}×400×(2.1-2)\)

\(=\dfrac{1}{2}×400×0.1\)

\(=20 \,N.s\)

The force versus time graph of a particle, moving along the x-axis is given. Calculate the impulse of the force during time t = 2 s to t = 2.1 s.

image
A

20 N.s

.

B

25 N.s

C

30 N.s

D

40 N.s

Option A is Correct

Impulse Momentum Theorem

  • In some situations, value of impulsive force acting on a particle, cannot be calculated, but its effect can be measured by calculating \(\displaystyle\int\overrightarrow{F}.dt\). This quantity is termed as IMPULSE.
  • So, mathematically

                  Impulse \((\overrightarrow{J})=\displaystyle\int\overrightarrow{F}.dt\)  ... (1)

                  From Newton's Second Law

                      \(\Delta\overrightarrow{P}=\displaystyle\int\overrightarrow{F}.dt\)             ... (2)

                    From (1) and (2)

                     \(\overrightarrow{J}=\Delta\overrightarrow{P}\)

  • The change in the momentum of a particle is equal to the impulse of the net force acting on the particle.
  • \(\overrightarrow{P}_f - \overrightarrow{P}_i=\overrightarrow{J}\)    ... (3)

  • This is an important result and is known as the Impulse- Momentum Theorem.

Illustration Questions

A ball, moving with a speed of 10 m/s collides with a wall and returns back with a speed of 5 m/s along the same line. Calculate the magnitude of impulse imparted by the wall. Mass of the ball is 2 kg.

A 50 N.s

B 12 N.s

C 25 N.s

D 30 N.s

×

image

Initial momentum =  \(\overrightarrow{P}_i =(2×10)\hat i\,kg\;m/s\;\)

Final momentum = \(\overrightarrow{P}_f =(2×5) (-\hat i)\,kg\;m/s\;\)

 

Change in momentum

\(\Delta\overrightarrow{P}=\overrightarrow{P}_f - \overrightarrow{P}_i\)

\(=[-10 \,\hat i - 20\,\hat i]\)

\(=30 (-\hat i)\,kg\;m/s\;\)     

By impulse momentum theorem

       \(\overrightarrow{J}=\Delta\overrightarrow{P}\)

      \(\overrightarrow{J}=30 (-\hat i)\,kg\;m/s\;\)

           \(=30(-\hat i)\,N.s\)

     \(|\overrightarrow{J}| =30\,N.s\)

A ball, moving with a speed of 10 m/s collides with a wall and returns back with a speed of 5 m/s along the same line. Calculate the magnitude of impulse imparted by the wall. Mass of the ball is 2 kg.

A

50 N.s

.

B

12 N.s

C

25 N.s

D

30 N.s

Option D is Correct

Calculation of Change in Velocity from F-T Graph

Calculation of Impulse from Force-Time Graph

  • We know that

\(\overrightarrow{J}=\displaystyle\int\limits_{t_i}^{t_f}\overrightarrow{F}.dt\)

  • It can be seen that area under F - t curve gives the magnitude of impulse.
  • By knowing the area under F - t curve, the change in momentum is known and hence, final speed can be calculated.

 

Illustration Questions

A ball of mass 2 kg is at rest at t = 0. If a force is applied as shown by F - t graph, then what will be it's final speed at t = 2 sec?

A 5 m/sec

B 10 m/sec

C 15 m/sec

D 12 m/sec

×

Calculation of area under the curve

\(=\dfrac{1}{2}×40×1\)

\(=20\,N.s\)

Let final momentum of ball is  \(m\overrightarrow{v}\).

Initial momentum = 0

\(\Delta \overrightarrow{P}=m\overrightarrow{v}-0\)

Area under the Force-time graph gives the change in momentum.

Area = \(\Delta \overrightarrow{P}\)

\(m\overrightarrow{v}-0=20\)

\(2×v - 0 =20\)

\(v=10\,m/sec\)

A ball of mass 2 kg is at rest at t = 0. If a force is applied as shown by F - t graph, then what will be it's final speed at t = 2 sec?

image
A

5 m/sec

.

B

10 m/sec

C

15 m/sec

D

12 m/sec

Option B is Correct

Practice Now