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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 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
A. $mng$		B. $mnv$
C. $mnvg$		D. $frac {mnv}{g}$
Right Option : B
View Explanation

Question : 2

A force of 10 N acts on a body of mass 20 kg for 10 seconds. Change in its momentum is
A. 5 kg m/s
B. 100 kg m/s
C. 200 kg m/s
D. 1000 kg m/s
Right Option : B
View Explanation

Question : 3

A ball of mass 40 g moving with a velocity of 5 $ms^-^1$ 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:
A. $4X10^8N$	B. $4X10^5N$	C. $4X10^2N$	D. $400N$
Right Option : B
View Explanation

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Newton Second Law in terms Of Momentum

Newton Second Law in terms Of Momentum

Students / Parents Reviews [10]

Archit Segal

Manish Kumar

Shreya Shrivastava

Hiya Gupta

Cheshta

Om Umang

Barkha Arora

Shivam Rana

Prisha Gupta

Eshan Arora