Change of momentum and its applications
Momentum is defined as a product
of mass and velocity. It is a physical quantity which explains that how much
kinetic energy is transferred between the two bodies during the collision. If a
body has higher momentum, it transfers more kinetic energy to the other bodies
and vice versa. Momentum is a vector quantity which has both magnitude and
direction. When we are calculating the change in the momentum of a body in
between any two cases, we shall use parallelogram law of vectors.
If a body is coming back with the
same velocity after striking an obstacle vertically, the momentum of the body
reminds the same in terms of the magnitude but the direction is opposite. Hence
the difference in the magnitude of the momentum is equal to double the initial
momentum of the system. It is simply because we need to calculate the
difference and hence we need to use a negative sign. We also need to use
another negative sign because the final momentum is opposite to the initial
momentum.
Application One
Let us consider a ball of mass M
moving with the velocity striking of all making an angle with the horizontal.
Let after striking the obstacle it bounce back with the same velocity and again
by making the same angle with the horizontal. In this case we can calculate the
change in the momentum as shown below.
Application Two
Let us consider another situation
where gravel is dropped vertically at at a constant rate on the conveyor belt
which is moving horizontally. As this gravel is acting agonist the motion of
the horizontal system, we need to apply a constant force to keep it moving. We
can get the value of that force simply by equating the force to the definition
of rate of change of momentum.
Application Three
Let us consider another case
where water is coming out of the nozzle through a horizontal hosepipe of a
certain area of cross-section. If the water is coming with a constant speed, in
the horizontal force on the pipe to prevent it to recoil can be calculated
using the definition of the force in terms of rate of change of momentum as
shown below.
Laugh conservation of linear momentum
If there is no external force
acting on a system, the linear momentum of the system always remains constant.
This is called conservation of linear momentum are simply momentum. Linear
momentum is conserved in all the cases if there are no external forces acting
on the system. This is one of the fundamental conserving physical quantities in
mechanics. Whatever may be the kind of the collision between the two bodies,
linear momentum is always conserved.
Impulse and its applications
Impulse is a physical quantity
which is defined as the product of force and time. It is actually a physical
quantity which explains how a large forces acting on a body for a very short
interval of time. This is in terms of magnitude is equal to the linear
momentum.
Example one
We can see the examples of the
impulsive force acting on a body in daily life at many situations. When the
cricket ball is hit by a bat, the batsmen hits the ball with a large force in a
very short interval of the time and hence a large impulsive forces acts on the
ball. We can see a small deformation in the shape of the ball also when the
ball is little bit weak to absorb as much of the force.
Example Two
When a fielder in the cricket
match tries to catch a ball, he pulls his hands back word in the process. This
is a common observation that any people who are familiar with the cricket know
it and see it. If he didn’t pull his hands backward, the ball hits his palms
with a larger force in a very short interval of time and it causes injury to
his palms. To avoid this he pulls the hands in the backward direction and hence
he is not allowing the entire large force acting on the body in a very short
interval of time, rather is giving the extra time for the ball to come to the
state of rest. In this process the force acting on the palm is distributed over
the time and per second that much of the large force is not acting on the palm
and hence he is not going to be injured.
Example Three
If two people are jumping from
the same height onto the ground, the person jumping onto the cement floor get
more injured than the person who is jumping from the same height but onto the
sand floor. The logic is simple. The person jumping onto the cement floor comes
to the state of rest immediately and hence the large impulsive force acts on
his body which causes injury. The person jumping onto the sand gets extra time
as the penetration little bit deep into the sand and hence he has less
possibility of getting that much injury.
Example Four
We also know that the shock
absorbers are used in all the motor vehicles to make the journey comfortable.
The shock absorbers do the same job and get more time for the force distribution
on the bodies of the people sitting in the vehicle and hence less impulsive
force will be acting on them. So the journey will be more comfortable.
Problem
A machine gun has a mass of 20
kg. It fires bullet of mass 35 gram at the rate of four bullets per second. The
bullet speed is 400 meter per second. What must be the force that has to be
applied to the gun to keep it on its position?
Solution
As there is no external force
acting on the system, according to laugh conservation of linear momentum
initial momentum of the system is equal to the final momentum. Therefore the
momentum of the gun is equal to the sum of the momentum of the bullets. By
equating that we can calculate the velocity of the recoil of the gun. We can
also equal to the momentum in terms of the magnitude to a physical quantity
called impulse.
By equating the momentum to the force
we can solve the problem as shown below.
Problem
A disc of March 0.1 kg is
floating horizontally in midair by firing the bullets of mass 0.05 kg each
vertically at it at the rate of 10 bullets per second. If the bullets rebound
with the same speed what is the speed with which each bullet has to be fired?
Solution
For the disc to be in the state
of equilibrium, the resultant force acting on it shall be equal to 0. The force acting in the downward
direction is the weight of the disc and the force acting in the upward
direction is due to the change in the momentum. If these two are equal in
magnitude as they were already in the opposite direction, the disc can be in
the state of equilibrium. Applying this concept we can solve the problem as
shown below.
Problem
The linear momentum of a particle
is varying with the time as shown below. What is the relation between the force
and time in this case?
Solution
We can define the force as rate
of change of momentum. By differentiating the given momentum once with respect to
the time hence we can get the force. Differentiation is a mathematical tool to
get the microscopic physical quantity from the macroscopic physical quantity.
By doing so we can get the relation and we can also draw a graph is as shown
below.
Problem
A liquid of known density is
passing through a horizontal pipe of uniform area of cross-section. The pipe is
a right angled and the liquid is travelling with a particular velocity. What
force has to be exerted at the bend to hold the pipe in the equilibrium
position?
Solution
We can calculate the change of
the momentum at the right angle using the parallelogram law of vectors. We can
also equate the change in the momentum to the force basing on the Newton’s
second law. By doing so, we can get the force as shown below.
Problem
A body of mass 10 kg is acted
upon by force which is varying with respect to the time as shown below. The
initial speed of the body is 10 m/s. What is the velocity of the body after
five seconds?
Solution
Until now we are dealing with a
constant force. But in this problem force is not constant but is varying with
respect to time. To solve this kind of the problems, we need to use a
mathematical tool called integration. In that case we can get the value of the
macro picture by knowing the value of the micro picture. It is performed as
shown in the below diagram.
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