Laws of motion
LAWS OF MOTION
Newton’s first law: Every body persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by external force on it.
From Newton’s first law, we get definition of force and inertia.
Force: Force is the push or pull which changes or tends to change the state of rest or of uniform motion in a straight line.
Inertia: This is the property by virtue of which a body tends to resist the change of its state of rest or of uniform motion in a straight line.
Inertia of rest: This is the property by virtue of which a body maintains its state of rest or tends to remain at rest.
Passengers sitting in a bus experience jerk in the backward direction when the bus suddenly starts moving. When the bus suddenly starts moving the lower part of the passengers share the motion but the upper part tends to remain at rest due to inertia of rest.
Inertia of motion: This is the property by virtue of which a body maintains its state of uniform motion or tends to maintain its motion.
Passengers standing in a moving bus experience jerk in the forward direction when the bus suddenly stops. When the bus suddenly stops the lower part of the passenger comes to rest but the upper part remains in the state of motion on account of inertia of motion.
Newton’s second law: The time rate of change of momentum of a body is directly proportional to the impressed force and takes place in the direction of the force.
Prove F = ma: Let us consider a body of mass m is moving with uniform velocity u.
Then it accelerates uniformly with a and after time t its final velocity is v.
The initial momentum of the body = mu and final momentum of the body = mv.
So, the change in momentum of the body = mv-mu.
According to Newton’s second law,
Applied force, F 
Or, F = 
(where k = constant)
Or, F = kma (v = u+at)
If one unit force is applied on a unit mass to produce unit acceleration, then k = 1. If k = 1, then F = ma.
Absolute unit of force:
C.G.S. unit is dyne (1 dyne = 1g 1cms-2). One dyne of force is the amount of force required to accelerate a body with 1 cms-2 of mass 1 gram.
SI unit is newton (1 newton = 1Kg 1ms-2). One newton of force is the amount of force required to accelerate a body with 1
ms-2 of mass 1Kg.
1N = 1Kg 1ms-2 or, 1N = 103 g 102cms-2 or, 1N = 105 gcms-2 or, 1N = 105 dyne.
Gravitational unit of force:
C.G.S. unit is gram-weight (gwt). One gram weight is the force with which a body of mass 1g is accelerated towards the centre of earth. 1gwt = 1g 981cms-2 = 981 dyne.
SI unit is kilogram-weight (kgwt). One kilogram weight is the force with which a body of mass 1Kg is accelerated towards the centre of earth. 1Kgwt = 1Kg 9.81ms-2 = 9.81 newton.
Dimension of force: F = ma, [F] = [MLT-2].
Impulse: The total effect of force is called impulse. It is the product of force and the time for which the force is applied.
Let us consider a body of mass m is moving with uniform velocity u. Then it accelerates uniformly with 
and after time t its final velocity is 
. The applied force,
= 
or, = .
The impulse 
= .t = = m(
– 
) = m – m = change of momentum of the body.
If the force is applied for time interval t1 to t2 then impulse = 
dt .
Impulsive force: When a heavy amount of force is applied on a body for a small interval of time to change of momentum of the body then that force is called impulsive force. Example: Collision between bat and ball.
Balance force: If the resultant force of all the forces acting on a body is zero, then those forces are called balance force.
Unbalance force: If the resultant force of all the forces acting on a body is nonzero, then those forces are called unbalance force. For unbalance force, resultant force = mass × acceleration.
Newton’s third law: To every action, there is always an equal (in magnitude) and opposite (in direction) reaction.
Law conservation of linear momentum: If the vector sum of the external forces acting on a system is zero, then the total momentum of the system is conserved i.e. remain constant.
Let us consider two bodies A and B of masses m1 and m2 are moving with initial velocity 
and 
( > ) in the same direction and after some time they collide and move in the same direction with final velocity 
and 
( < ) respectively.
The change in momentum of the body A = m1( – ) and the change in momentum of the body B = m2( – ).
If the collision time is t, and 
is the action of B on A and 
is the action of A on B, then from Newton’s third law,
= –
Or,t = – t —–(i).
The impulse of body A =t = m1( – )and impulse of body B = t = m2( – ).
Then from equation (ii) we get, m1( – ) = – m2( – )
Or, m1 – m1 = – m2 + m2
Or, m1+ m2 = m1 + m2 .
So, the momentum of the system before collision = the momentum of the system after collision.
Recoil of gun: Let us consider, initially the gun and bullet are rest. So, the momentum of the system before firing is zero.
Let m = mass of bullet, M = mass of gun, = velocity of bullet, 
= velocity of gun after firing. As there is no external force is applied then the total momentum of the system is conserved.
So, from conservation of linear momentum, 0 = m +M
Or, = – 
[– ve sign indicates that the motion of gun is opposite to that of bullet.]
Concepts to solve numerical based on Newton’s law:
(i) Consider a system on which laws of motion are to be applied. It may be a body or a point mass or a single block or the combination of blocks.
(ii) After identifying the system, consider all the forces acting on the system. Ignore the forces which are applied by the system on other body.
(iii) If the forces are coplanar then resolute all the forces along x axis and y axis.
(iv) If the body is in equilibrium, then 
= 
= 0. If the body is in motion, then = m
and = m
.
Concepts on Newton’s law:
- A ball of mass m is thrown perpendicularly on the wall with speed u and rebounces with speed v. The collision time of ball with wall is t. Find force on the ball by wall.
The momentum of the ball before collision with the wall is Pi =- mu [taking the direction of momentum along + ve x axis is positive.]
The momentum of the ball after collision with the wall is Pf = mv
Change of momentum P = Pf – Pi = mv – (- mu) = m(u+v)
Force on the ball by wall F =
[direction is along +ve x axis].
- A ball of mass m is thrown on the wall with speed u at an angle and rebounces with speed v at same angle as shown in figure. The collision time of ball with wall is t. Find force on the ball by wall.
Let us consider the direction of momentum along + ve x axis and + ve y axis are positive.
The momentum of the ball before collision with the wall is Pi = mu
Andresoluting it along x and y axis we get, Pix = – mu sin
and Piy = – mu cos .
The momentum of the ball after collision with the wall is Pf = mv and resoluting it along x and y axis we get, Pfx = mv sin and Pfy = – mv cos .
Change in momentum along x axis is Px = Pfx – Pix = mv sin – (- mu sin ) = m sin (u+v)
Change in momentum along y axis is Py = Pfy – Piy = – mvcos -(- mucos ) = mcos (u-v)
Net change in momentum P =
= 
Therefore, force on the ball by wall F 
= [direction is along +ve x axis].
- A stream of water flowing horizontally with a speed u gushes out of a tube of cross-sectional area A and hits a vertical wall nearby. Find the force exerted on the wall by the impact of water. [Assume water does not rebound]
Let us consider m mass of water hits the wall for time t.
So, m = volume of water flows for time t density of water
m = Area × distance travelled by water for time t × density of water
m = Aut
[density of water is ]
Momentum of water before collision with the wall [direction perpendicular to the wall] = mu = Au2 t
As the water does not rebound so after collision with wall, momentum of water is zero [mv = mucos900 = 0].
Therefore, force exerted on the wall by the impact of water is
F = 
= 
= 
.
- Concepts on free body diagram: (i) A uniform rope of length L, resting on a frictionless horizontal surface is pulled at one end by a force F. Find the tension in the rope at a distance l from the end where the force is applied.
Let us consider the mass of the rope is M and mass per unit length of the rope is 
.
So, the mass of the part of length l is m =
. Acceleration of rope is a = 
.
Considering the part of the rope of length l as a system the forces act on it are F and T
(T is tension applied on this part by the other part).
Then using Newton’s 2ndlaw we get, F -T = ma = 
∴ T = F – = 
.
(ii) Two bodies of mass M and m are connected by string and m mass is pulled by a force F on frictionless surface. Find tension acting along the string.
Consider two bodies as a system and F is the external force acting on the system.
So, the acceleration of the system is a = 
.
Now consider the body of mass m as a system and F and T (tension) act on the body.
The net force acts on it is 
= F -T = ma [as the acceleration is directed along the direction of external force F, so F > T]
Or, T = F – ma = F – 
= 
.
If we consider the body of mass M as a system and T is the external force acts on it by mass m then T = Ma = 
.
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