Newton Second Law in terms Of Momentum


 
 
Concept Explanation
 

Newton Second Law in terms Of Momentum

Newton’s Second Law in terms of Momentum

In terms of momentum second law can be stated as:

The rate of change of momentum of an object is proportional to the net force applied on the object. The direction of the change of momentum is the same as the direction of the net  force.

Let us see how the two ways of stating the second law are equivalent. Suppose a object of mass m is moving along a straight line with a constant acceleration a. Also, suppose its velocity at time  t_1 is v_1, which changes to  v_2  at time  t_2.

The linear momentum at time t_1 is  p_1:=:mv_1.

At time t_2 is p_2:=:mv_2

Here, p = momentum ,t= time  and  v = velocity

The rate of change of momentum is the ratio of change in momentum with change in time

  =frac{p_2-p_1}{t_2-t_1}   

According to the second law, it is proportional to force.

                                                         frac{p_2-p_1}{t_2-t_1}:alpha :F

    or                                                  F:=:kfrac{p_2-p_1}{t_2-t_1}          (where k is a constant)     

   or                                                  F:=:kfrac{mv_2-mv_1}{t_2-t_1}:=:km[frac{v_2-v_1}{t_2-t_1}]

  or                                                   F:=:kma                            ....................(ii)

Equation (ii) is same as that of equaiton (i)

Q 1. A 650 kg rocket isto be speeded up from 440 metres per second to 520 meters per second in outer space. If the thrust of the engine is 1200 N, for how long must the engine be fired?

Solution:   The change in the momentum of the rocket    Delta p:=:mv:-:mu            

                           = (650) (520) - (650) (440)  =  52000 kg m/s.

                  This must be equal to the impulse, so FDelta t = (1200 N) (Delta t)  =  52000 kg m/s

                  therefore     Delta t  =   43 s

Q 2.  A boy of mass 58 kg jumps with a horizontal velocity of 3 m/s onto a stationary skateboard of mass 2 kg. What is his velocity as he moves off the skateboard ?

Solution: Assume there is zero unbalanced horizontal force in the horizontal direction and that left to right is the positive (+) direction.

                Momentum before interaction  =   m_1u_1+m_2u_2:=58times3:+;2times0

                Momentum  after interaction  =  m_1v_1:+m_2v_2:=:58times:v:+:2times:v:=:60v

              Equating   momenta                  60v:=:58times:3

                                                                v:=:frac{58times3}{60}:=:+ 2.9: ms^{-1}

                                                                                                 

   

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

A ball of mass 40 g moving with a velocity of 5ms^-^1 strikes a wall normally and rebounds with the same velocity. The time of contact between the ball and wall is 1micro second. Ther average force exerted by the wall on the ball is:

Right Option : B
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Explanation
Question : 2

A Diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 ms^-^1. The accelerating force on the rocket is

Right Option : B
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Explanation
Question : 3

A machine gun fires n bullets per second and the mass of each bullet is m. It the speed of bullets is v, then the force exerted on the machine gun is

Right Option : B
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Explanation
 
 
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