WBCHSE Class 11 Physics Impulse Of Force And Impulsive Force Notes

Newton Law Of Motion – Impulse Of Force And Impulsive Force

Impulse Of Force: When a cricket ball is hit by a bat or a nail is struck into wood with a hammer, the force of impact

  1. Acts for a short time, and
  2. Is usually not constant throughout the duration of impact i.e., It varies with time.

It is not easy to measure the varying force of impact. Let \(\vec{F}_{a v}\) be the average force acting during the small time of impact \(\vec{F}_{a v}\) t. The quantity \(\vec{F}_{a v}\)t is called the impulse of a force.

Impulse Of Force And Impulsive Force Definition: For a force acting on a body for an interval of time, the product of the force and the time interval is called the impulse of the force or simply impulse.

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Let a force \(\vec{F}\) act for small time dt. The impulse of the force is given by d\(\vec{I}\) = \(\vec{F}\)dt

If we consider a finite interval of time from t1 to t2, then the impulse will be, \(\vec{I}=\int d \vec{I}=\int_{t_1}^{t_2} \vec{F} d t\)

If \(\vec{F}_{a v}\) is the average force, then, \(\vec{I}=\vec{F}_{a v}\left(t_2-t_1\right)\)

or, \(\vec{I}=\vec{F}_{a v} \Delta t \text {, where } \Delta t=t_2-t_1 \text {. }\)

Impulse-Momentum Theorem: According to Newton’s second law of motion, applied force = rate of change in momentum

∴ \(\vec{F}=\frac{d \vec{p}}{d t} \quad \text { or, } \vec{F} d t=d \vec{p}\)

If in time 0 to t, the momentum of the body changes from p1 to p2, then integrating within proper limits, we get, \(\int_0^t \vec{F} d t=\int_{\vec{p}_1}^{\vec{p}_2} d \vec{p}=d \vec{p}=\left(\vec{p}_2-\vec{p}_1\right)\)

But \(\int_0^t \vec{F} d t=\vec{I} \quad therefore \vec{I}=\left(\vec{p}_2-\vec{p}_1\right)\)

Thus, the impulse of a force is equal to the total change in momentum produced by the force. This relationship between impulse and momentum is known as the impulse-momentum theorem.

Unit And Dimension Of Impulse: impulse has the unit and dimension of momentum:

  • CGS System: g · cm · s-1  or dyn · s
  • SI: kg · m · s-1 or N · s

The dimension of impulse is MLT-1.

Measurement Of Impulse By Graphical Method:

1. When A Constant Force Force Acts On A Body: Let a constant force F act on a body from time t1 to t2. The force-time graph is a straight line AB parallel to the time axis, as shown.

Newtons Law Of Motion Measurment Of Impulse By Graphical Method

Impulse of the constant force,

l = area of rectangle ABCD

= ADxAB = F(t2 – t1)

2. When A Variable Force Acts On The Body: Suppose a force varying in magnitude acts on a body for time t2 – t1 = t

The force-time graph is a curve ABC as shown.

Newtons Law Of Motion Variable Force Acts On A Body

Impulse of force F in time interval t, I = \(\int_0^t F d t\) = area under the curve ABC

Thus, the area under the force-time graph gives the magnitude of the impulse of the given force in the given time interval.

Practical Applications Of Impulse: If two forces \(\vec{F}_1\) and \(\vec{F}_2\) act on a body to produce the same impulse (or change in momentum), then their time durations t1 and t2 should be such that, \(\vec{F}_1\)t1 = \(\vec{F}_2\)t2

Newtons Law Of Motion Practical Applications Of Impulse

In other words, if the time duration of an impulse is large, the force exerted will be small, or vice-versa. The Curves (1) and (2) indicate forces applied and the time intervals for which they act are different but the area under the F-t curve is the same implying that the same impulse is produced in both cases.

The following examples will make the concept clear.

  1. While catching a ball, a cricket player lowers his hands to save himself from getting hurt: By lowering his hands, the cricket player increases the time interval in which the catch is completed. As the total change in momentum takes place in a large time interval, the time rate of change of momentum of the ball decreases. So, according to Newton’s second law of motion, a lesser force acts on the hands of the player which saves him from getting hurt.
  2. A person falling from a certain height receives more injuries if he lands on a cemented floor than on soft ground (or loose earth, soft snow, cotton, or net). On the cemented floor, the momentum is reduced to zero in comparatively less time. Due to this, the rate of change of momentum is large. So, greater force acts on the person resulting in more injuries.
  3. Automobiles (buses, cars, etc) are provided with shockers. When a vehicle moves on an uneven road, it experiences jerks. The shocker increases the time of jerk and hence reduces the force.
  4. Chinawares are wrapped in straw or paper before packing. The straw (or, paper) between the chinawares increases the time of experiencing the jerk during transportation. Hence, they strike against each other with a lesser force and are less likely to be damaged.

Newton Law Of Motion – Impulse Of Force Numerical Examples

Example 1. The graph ABCD represents the change in force (N) against time (μs). Find the impulse on the body between 4 μs to 16 μs.

Newtons Law Of Motion A Graph ABCD Represents Impulse Of Body

Solution:

The impulse of the external force during 4 μs to 16 μs

= the area of the quadrilateral BCDE

= sum of the areas of the trapezium BCFE and triangle CDF

= 1/2(BE+ CF) x EF+ x CF = 1/2(200 + 800) x (6 – 4) + 1/2(16 – 6) x 800

= 5000 N · μs = 0.005 N · s [1 μs = 10-6 s]

Example 2. The initial speed of a body of mass 2.0 kg is 5.0 m/s. A force acts for 4 s in the direction of motion of the body. The force-time graph is shown. Calculate the impulse of the force and the final speed of the body.

Newtons Law Of Motion Intial Speed Of Body Of Mass Graph Representation

Solution:

Impulse of the force

= area between the force-time graph and the time axis

= area of triangle OAA’ + area of rectangle

AA’B’B + area of trapezium BB’C’C+ area of rectangle CC’D’D

= 1/2 = x 1.5 x 3 + 1 x 3 + 1/2(3 + 2)(3- 2.5) + 2 x 1

= 2.25 + 3 + 1.25 + 2 = 8.5 N · s

∴ Impulse = change in momentum = mΔv

∴ Change in velocity, \(\Delta v=\frac{\text { impulse }}{\text { mass }}\)

= \(\frac{8.5}{2}=4.25 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Final speed of the body= initial speed + Δv

= (5.0 + 4.25) = 9.25 m · s-1

Impulsive Force Definition: If a large force acts on a body for a very short interval of time, it is called an impulsive force.

Impulsive Force Example:

  1. When a nail is hammered, the applied force, though large, acts for a very short period of time. Hence it is an impulsive force. This force sets up an impulse on the nail and the nail penetrates the wall.
  2. A football is kicked with a large force, but the ball and the foot remain in contact for a very short time. This is an impulsive force acting on tKfe ball.
  3. In cricket, an impulsive force is applied on the ball by the bat when a batsman strikes the ball and the ball gets an impulse.
  4. In a game of carrom, the player applies an impulsive force on the striker and the striker also exerts an impulsive force on the pieces.
  5. When a man jumps from a height onto the ground, the ground exerts an impulsive force on the man and he quickly comes to rest.

Differences Between Impulse Of A Force And Impulsive Force:

Newtons Law Of Motion Differences Between Impulse Force And Impulsive Force

Newton Law Of Motion – Impulsive Force Numerical Examples

Example 1. A hammer of mass 1 kg hits a nail at a speed of 10 m s-1. For this, the nail penetrates 2 cm through a wooden plank. Calculate

  1. Impulse due to the hammer,
  2. The applied force and
  3. The time of contact between the hammer and the nail.

Solution:

Impulse due to the hammer

= change in momentum of the hammer

= m(v-u) = 1 x (10-0) = 10 kg · m · s-1

The nail penetrates 2 cm through the wooden plank. Let the time for which the hammer is in contact with the nail be t s.

∴ Distance moved = average velocity x time

or, \(\frac{2}{100}=\frac{10+0}{2}\) x t (average velocity = \(\frac{u+v}{2}=\frac{10+0}{2}\))

or, \(t=4 \times 10^{-3} \mathrm{~s}\)

If the applied force is F, then F · t=10

or, F = \(\frac{10}{4 \times 10^{-3}}=\frac{10000}{4}=2500 \mathrm{~N}\).

Example 2. Water, ejected from the jet of a firefighting engine, hits a wall perpendicularly at the rate of 12.2 m · s-1. Assuming that the water does not recoil from the wall, calculate the pressure developed on the wall. Mass of 1 m3 of water 103 kg.
Solution:

Water, ejected from the jet of a firefighting engine, hits a wall perpendicularly at the rate of 12.2 m · s-1.

Let the area of the cross-section of the jet be A.

Hence, the volume of water ejected per second = Axv and mass of water hitting the wall per second =A x v x d; d = density of water = 103 kg · m-3.

Change of momentum, per second, of the water jet hitting the wall

= Avdx (v – 0) = Av²d = A x 12.2 x 12.2 x 103 which is the force on the wall.

Hence, pressure on the wall = \(\frac{\text { force }}{\text { area }}=\frac{A \times 12.2 \times 12.2 \times 10^3}{A}\)

= 1.49 x 105 N · m-2.

Example 3. Two bodies of equal mass are at rest, side by side. A constant force F is applied on the first body while at the same instant an impulsive force, producing an impulse I is applied in the same direction on the second body. Determine the time taken by them, in terms of F and I, to be side by side again.
Solution:

Two bodies of equal mass are at rest, side by side. A constant force F is applied on the first body while at the same instant an impulsive force, producing an impulse I is applied in the same direction on the second body.

Let the mass of each body be m, and the times when they are side by side be 0 and t s.

Acceleration of the first body due to F, a = \(\frac{F}{m}\).

∴ Displacement of the first body in time t, starting from rest, \(s_1=\frac{1}{2} \cdot \frac{F}{m} \cdot t^2=\frac{F t^2}{2 m}\)…(1)

The change in momentum of the second body due to impulse,

I = change in momentum = mv- mu = mv-mx 0 = mv

or, v = \(\frac{I}{m}\)

As no other force acts on the second body, it moves with this constant velocity (v) for t s, and covers a distance s2

∴ \(s_2=v t=\frac{I t}{m}\)…(2)

According to the problem s1 = s2.

∴ \(\frac{F t^2}{2 m}=\frac{I t}{m}\) (from (1) and (2))

∴ t = \(\frac{2 I}{F}\).

Example 4. A cricket ball of mass 150 g, moving at 12 m · s-1, is hit by a cricket bat. The ball recoils with a velocity of 20 m · s-1. If the bat was in contact with the ball for 0.01 s, find the average force imparted on the ball by the bat.
Solution:

A cricket ball of mass 150 g, moving at 12 m · s-1, is hit by a cricket bat. The ball recoils with a velocity of 20 m · s-1. If the bat was in contact with the ball for 0.01 s

The change in velocity of the ball = {12-(-20)} = 32 m · s-1.

Hence, impulse or change in momentum

= \(\frac{150}{1000} \times 32 \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}\)

Average force x time of impact = impulse

If F is the average force, \(F \cdot t=\frac{150 \times 32}{1000}\)

or, \(F=\frac{150 \times 32}{1000 \times 0.01} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-2}=480 \mathrm{~N} .\)

Example 5. A person of mass 60 kg jumps from a height of 5 m, onto the ground. If he does not bend his knees on touching the ground, he comes to rest in 1/10s. But if he bends his knees, he takes 1 s to come to rest. Find the force exerted by the ground on him in the two cases, [g = 10 m · s-2]
Solution:

A person of mass 60 kg jumps from a height of 5 m, onto the ground. If he does not bend his knees on touching the ground, he comes to rest in 1/10s. But if he bends his knees, he takes 1 s to come to rest.

Let v = velocity acquired during the free fall through 5 m.

∴ v2² = 0 + 2 x 10 x 5 or, v= 10 m· s-1

This velocity becomes zero due to the impact with the ground.

Then, change in momentum = mv- mu = 60 x 10 – 0 = 600 N · s-1

The force applied by the earth for this change,

In 1st case, \(F_1=\frac{600}{\frac{1}{10}}=6000 \mathrm{~N}\)

In 2nd case, \(F_2=\frac{600}{1}=600 \mathrm{~N}\)

Hence, the impact on the case is more severe than in the 2nd case, where the momentum changed at a slower rate.

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