- Linear Momentum
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- Impulse and Momentum
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- Conservation of Linear momentum
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- Recoil of the Gun
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- Motion of the system with varying mass(Rocket)
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- Linear Momentum Problems with Solutions

- To explain the terms impulse and momentum consider a particle of mass m is moving along x-axis under the action of constant force F as shown below in the figure

- If at time t=0 ,velocity of the particle is v
_{0}then at any time t velocity of particle is given by the equation

**v**=**v**_{0}+**a**t

where**a**=**F**/m

can be determined from the newton's second law of motion .Putting value of acceleration in above equation\

we get

m**v**= m**v**_{0}+**F**t

or

**F**t = m**v**- m**v**_{0}-(1)

- right side of the equation Ft, is the product of force and the time during which the force acts and is known as the impulse

Thus

Impulse=**F**t

- If a constant force acts on a body during a time from t
_{1}and t_{2},then impulse of the force is

**I**=**F**(t_{2}-t_{1}) -(2)

Thus impulse received during an impact is defined as the product of the force and time interval during which it acts

- Again consider left hand side of the equation (1) which is the difference of the product of mass and velocity of the particle at two different times t=0 and t=t

- This product of mass and velocity is known as
**linear momentum**and is represented by the symbol**p**. Mathematically

**p**= m**v**--(3)

- physically equation (1) states that the impulse of force from time t=0 to t=t is equal to the change in linear momentum during
- If at time t
_{1}velocity of the particle is v_{1}and at time t_{2}velocity of the particle is v_{2},then

F(t_{2}-t_{1})=mv_{2}-mv_{1}-(4)

- so far we have considered the case of the particle moving in a straight line i.e along x-axis and quantities involved F,v, and a were all scalars

- If we call these quantities as components of the vectors
**F**,**v**and**a**along x-axis and generalize the definitions of momentum and impulse so that the motion now is not constrained along one -direction ,Thus we got

Impulse=**I**=**F**(t_{2}-t_{1}) -(5)

Linear momentum=**p**=m**v**-(6)

where

**I**=I_{x}**i**+I_{y}**j**+I_{z}**k**

**F**=F_{x}**i**+F_{y}**j**+F_{z}**k**

**p**=p_{x}**i**+p_{y}**j**+p_{z}**k**

**v**=v_{x}**i**+v_{y}**j**+v_{z}**k**

are expressed in terms of their components along x,y and z axis and also in terms of unit vectors - On generalizing equation (4) using respective vectors quantities we get the equation

**F**(t_{2}-t_{1}) =m**v**_{2}-m**v**_{1}-(7) - So far while discussing Impulse and momentum we have considered force acting on particle is constant in direction and magnitude

- In general ,the magnitude of the force may vary with time or both the direction and magnitude may vary with time

- Consider a particle of mass m moving in a three-dimensional space and is acted upon by the varying resultant force
**F**. Now from newtons second law of motion we know that

**F**=m(d**v**/dt)

or**F**dt=md**v**

- If at time t
_{1}velocity of the particle is**v**_{1}and at time t_{2}velocity of the particle is**v**_{2},then from above equation we have

- Integral on the left hand side of the equation (8) is the impulse of the force
**F**in the time interval (t_{2}-t_{1}) and is a vector quantity,Thus

Above integral can be calculated easily if the Force F is some known function of time t i.e.,

F=F(t)

- Integral on the right side is when evaluated gives the product of the mass of the particle and change in the velocity of the particle

- using equation (9) and (10) to rewrite the equation (8) we get

- Equivalent equations of equation (11) for particle moving in space are

- Thus we conclude that impulse of force
**F**during the time interval t_{2}-t_{1}is equal to the change in the linear momentum of the body on which its acts - SI units of impulse is Ns or Kgms
^{-1}

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 5 m/s If the mass of the ball is 0.20 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)

Impulse =change in momentum

=.20 * 5 - (-.20* 5)

=2 N s

Class 11 Maths Class 11 Physics Class 11 Chemistry