Second Law of Motion
The first
law of motion indicates that when an unbalanced external force acts on an
object, its velocity changes, that is, the object gets an acceleration. We
would now like to study how the acceleration of an object depends on the force
applied to it and how we measure a force. Let us recount some observations from
our everyday life. During the game of table tennis if the ball hits a player it
does not hurt him. On the other hand, when a fast moving cricket ball hits a
spectator, it may hurt him. A truck at rest does not require any attention when
parked along a roadside. But a moving truck, even at speeds as low as 5 ms–1,
may kill a person standing in its path. A small mass, such as a bullet may kill
a person when fired from a gun. These observations suggest that the impact
produced by the objects depends on their mass and velocity. Similarly, if an
object is to be accelerated, we know that a greater force is required to give a
greater velocity. In other words, there appears to exist some quantity of
importance that combines the object’s mass and its velocity. One such property
called momentum was introduced by Newton. The momentum, p of an object is
defined as the product of its mass, m and velocity, v.That is,
p = mν(1)
Momentum
has both direction and magnitude. Its direction is the same as that of
velocity, ν. The SI unit of momentum is kilogram-meter per second (kg m s-1).
Since the application of an unbalanced force brings a change in the velocity of
the object, it is therefore clear that a force also produces a change of
momentum.
Let us
consider a situation in which a car with a dead battery is to be pushed along a
straight road to give it a speed of 1 m s-1, which is sufficient to
start its engine. If one or two persons give a sudden push (unbalanced force)
to it, it hardly starts. But a continuous push over some time results in a
gradual acceleration of the car to this speed. It means that the change of
momentum of the car is not only determined by the magnitude of the force but
also by the time during which the force is exerted. It may then also be
concluded that the force necessary to change the momentum of an object depends
on the time rate at which the momentum is changed.
The
second law of motion states that the rate of change of momentum of an object is
proportional to the applied unbalanced force in the direction of force.
MATHEMATICAL
FORMULATION OF SECOND LAW OF MOTION
Suppose
an object of mass, m is moving along a straight line with an initial velocity,
u. It is uniformly accelerated to velocity, ν in time, t by the application of
a constant force, F throughout the time, t. The initial and final momentum of
the object will be, p1 = mu and p2 = mν respectively.
The
change in momentum
α p2 – p1
α mν – mu
α m × (ν – u).
The rate
of change of momentum α m × (ν −u) / t
Or, the
applied force, F α m × (ν −u) / t
Or, the
applied force, F = km × (ν −u) / t(2)
= kma(3)
Here a [
= (ν-u) / t] is the acceleration, which is the rate of change of velocity. The
quantity, k is a constant of proportionality. The SI units of mass and
acceleration are kg and m s-2 respectively. The unit of force is so
chosen that the value of the constant, k becomes one. For this, one unit of
force is defined as the amount that produces an acceleration of 1 m s-2
in an object of 1 kg mass. That is,
1 unit of
force = k × (1 kg) × (1 m s-2).
Thus, the
value of k becomes 1. From Eq. (3)
F = ma(4)
The unit
of force is kg m s-2 or newton, which has the symbol N. The second
law of motion gives us a method to measure the force acting on an object as a
product of its mass and acceleration.
The
second law of motion is often seen in action in our everyday life. Have you
noticed that while catching a fast moving cricket ball, a fielder in the ground
gradually pulls his hands backwards with the moving ball? In doing so, the
fielder increases the time during which the high velocity of the moving ball
decreases to zero. Thus, the acceleration of the ball is decreased and
therefore the impact of catching the fast moving ball (Fig. 8) is also reduced.
If the ball is stopped suddenly then its high velocity decreases to zero in a
very short interval of time. Thus, the rate of change of momentum of the ball
will be large. Therefore, a large force would have to be applied for holding
the catch that may hurt the palm of the fielder. In a high jump athletic event,
the athletes are made to fall either on a cushioned bed or on a sand bed. This
is to increase the time of the athlete’s fall to stop after making the jump.
This decreases the rate of change of momentum and hence the force. Try to
ponder how a karate player breaks a slab of ice with a single blow.
fig. 8 A fielder pulls his hands
gradually with the moving ball while holding a catch.
The first
law of motion can be mathematically stated from the mathematical expression for
the second law of motion. Eq. (4) is
F = ma
or F = m(ν-u)/t(5)
or Ft =
mν - mu
That is,
when F = 0, ν = u for whatever time, t is taken. This means that the object
will continue moving with uniform velocity, u throughout the time, t. If u is
zero then ν will also be zero. That is, the object will remain at rest.
Third Law of
Motion
The first
two laws of motion tell us how an applied force changes the motion and provide
us with a method of determining the force. The third law of motion states that
when one object exerts a force on another object, the second object
instantaneously exerts a force back on the first. These two forces are always
equal in magnitude but opposite in direction. These forces act on different
objects and never on the same object. In the game of football sometimes we,
while looking at the football and trying to kick it with a greater force,
collide with a player of the opposite team. Both feel hurt because each applies
a force to the other. In other words, there is a pair of forces and not just
one force. The two opposing forces are also known as action and reaction
forces.
Let us
consider two spring balances connected together as shown in Fig. 10. The fixed
end of balance B is attached with a rigid support, like a wall. When a force is
applied through the free end of spring balance A, it is observed that both the
spring balances show the same readings on their scales. It means that the force
exerted by spring balance A on balance B is equal but opposite in direction to
the force exerted by the balance B on balance A. The force which balance A
exerts on balance B is called the action and the force of balance B on balance
A is called the reaction. This gives us an alternative statement of the third
law of motion i.e., to every action there is an equal and opposite reaction.
However, it must be remembered that the action and reaction always act on two
different objects.
Fig.10 Action and reaction forces are
equal and opposite.
Suppose
you are standing at rest and intend to start walking on a road. You must
accelerate, and this requires a force in accordance with the second law of
motion. Which is this force? Is it the muscular effort you exert on the road?
Is it in the direction we intend to move? No, you push the road below
backwards. The road exerts an equal and opposite reaction force on your feet to
make you move forward.
It is
important to note that even though the action and reaction forces are always
equal in magnitude, these forces may not produce accelerations of equal
magnitudes. This is because each force acts on a different object that may have
a different mass.
When a
gun is fired, it exerts a forward force on the bullet. The bullet exerts an
equal and opposite reaction force on the gun. This results in the recoil of the
gun (Fig. 11).Since the gun has a much greater mass than the bullet, the
acceleration of the gun is much less than the acceleration of the bullet. The
third law of motion can also be illustrated when a sailor jumps out of a rowing
boat. As the sailor jumps forward, the force on the boat moves it backwards
(Fig. 12).
Fig.11 A forward force on the bullet
and recoil of the gun.
Fig.12 As the sailor jumps in forward
direction, the boat moves backwards.
Activity
4
- Request two children to
stand on two separate carts as shown in Fig. 13.
- Give them a bag full of sand
or some other heavy object. Ask them to play a game of catch with the bag.
- Does each of them receive an
instantaneous reaction as a result of throwing the sand bag (action)?
- You can paint a white line
on cartwheels to observe the motion of the two carts when the children
throw the bag towards each other.
Fig.13
Now,
place two children on one cart and one on another cart. The second law of
motion can be seen, as this arrangement would show different accelerations for
the same force.
The cart
shown in this activity can be constructed by using a 12 mm or 18 mm thick
plywood board of about 50 cm × 100 cm with two pairs of hard ball-bearing
wheels (skate wheels are good to use). Skateboards are not as effective because
it is difficult to maintain straight-line motion.
Conservation
of Momentum
Suppose
two objects (two balls A and B, say) of masses mA and mB
are traveling in the same direction along a straight line at different
velocities uA and uB, respectively [Fig. 14(a)]. And
there are no other external unbalanced forces acting on them. Let uA
> uB and the two balls collide with each other as shown in Fig.
14(b). During collision which lasts for a time t, the ball A exerts a force FAB
on ball B and the ball B exerts a force FBA on ball A. Suppose vA
and vB are the velocities of the two balls A and B after the
collision, respectively [Fig. 14(c)].
Fig.14 Conservation of momentum in
collision of two balls.
From Eq.
(1), the momenta (plural of momentum) of ball A before and after the collision
are mAuA and mAvA, respectively.
The rate of change of its momentum (or FAB, action) during the
collision will be mA (vA-uA) / t
Similarly,
the rate of change of momentum of ball B (= FBA or reaction) during
the collision will be mB (vB-uB) / t
According
to the third law of motion, the force FAB exerted by ball A on ball
B (action) and the force FBA exerted by the ball B on ball A
(reaction) must be equal and opposite to each other. Therefore,
FAB = – FBA(6)
or, mA (vA-uA)/t = - mB (vB-uB)/t
This
gives,
mAuA + mBuB = mAvA
+ mBvB(7)
Since (mAuA
+ mBuB) is the total momentum of the two balls A and B
before the collision and (mAvA + mBvB)
is their total momentum after the collision, from Eq. (7) we observe that the
total momentum of the two balls remains unchanged or conserved provided no
other external force acts.
As a
result of this ideal collision experiment, we say that the sum of momenta of
the two objects before collision is equal to the sum of momenta after the
collision provided there is no external unbalanced force acting on them. This
is known as the law of conservation of momentum. This statement can
alternatively be given as the total momentum of the two objects is unchanged or
conserved by the collision.
No comments:
Post a Comment