When
a person fire with a gun, he feels a backward push at that time. What is the
reason
behind
this? In this case, action and reaction force of bullet and gun exists for the
same
time.
According to the Newton’s second law of motion, the bullet and the gun acquire
equal
and opposite momentum. As a result the bullet moves forward with a particular
momentum
and the gun moves backward with a momentum of same magnitude but
opposite
in direction. Due to this, the person will feel a backward thrust. The backward
velocity
of the gun will be smaller in comparison to that of the bullet as the mass of
the
gun
is large.
Collision
Among
you, those who have played marble probably had seen how one marble hits
another
marble. In addition, you are familiar with different types of road accident
through
newspapers or television. These events are the practical examples of collision.
Therefore,
when a moving object hits another body at rest or in motion, then it is said
that
a collision has taken place. During collision a force acts on each of the two
bodies.
If
the force exerted by the first body on the second body is called action force,
then the
force
exerted by the second body on the first one is called the reaction force. These
two
forces
acting during collision are same in magnitude but opposite in direction. No
other
external
force acts during the collision except the action and reaction force. From
Newton’s
second law we get,
F mv mu
t
−
=
We
can express the change of momentum from this equation as-
F ×t
= mv −mu
(3.4)
i.e.
force × time = change of momentum.
The
product of force and time is defined as impulse of force.
Therefore,
impulse of force = change of momentum
Let
two bodies A and B having masses m1
and m2
are moving with velocities u1 and u2
respectively
along a straight line. If the velocity of A is greater than that of B, at any
time
the object A will collide with the object B [Figure 3.7]. The force exerted on
B by
A
is the action F1.
The object B will also exert a force F2
on A, this F2 is the reaction
force.
According to Newton’s third law of motion, 1
2 F = −F
Figure:
3.7
Physics
59
During
collision, the action and reaction force exists for the same time. Let the time
duration
of action and reaction be t. After the collision the two objects will continue to
move
along the same straight line with their changed velocities. Let v1 and v2 be the
changed
velocities of A and B respectively. If due to action and reaction, the
accelerations
of A and B are a1 and a2 respectively, then
F1 = −F2
or,
1 1 2 2 m a = −m
a
or,
1 1 2 2
1
2
m v u m v u
t t
− −
= −
or,
1 1 1 1 2 2 2 2 m v −mu
= −m v + m
u
or,
1 1 2 2 1 1 2 2 m u + m
u = m v + m
v
Therefore,
the sum of the momentum of the objects A and B remains same before and
after
the collision. It is the law of conservation of momentum.
Mathematical Example 3.3: A force of 2000 N acts on a body of mass 20 kg for a time
of
0.1 s. What is the change of momentum of the body?
Solution:
We know,
change of
momentum = force × time
Here,
mv −mu
= Ft applied force, F
= 2000 N
=2000
N × 0.1
s time duration, t =0.1 s
=
200 kg ms-2 s
change of momentum, mv
-mu =?
=200
kg ms-1
Ans: change of momentum = 200 kg ms-1
Mathematical Example 3.4: A bullet of mass 10 g was shot from a gun with a velocity
of
500 ms-1.
If the mass of the gun is 2 kg, find the backward velocity of the gun.
Solution:
Let
the direction of the bullet’s velocity i.e. the forward direction be positive.
From the
conservation
of momentum,
We
get,
1
1 2 2 1 1 2 2 m u + m
u = m v +m
v
ev
m1×
0 ms-1 +
m2 kg
× 0 ms-1=
10-2 kg
× 500 ms-1 +
2
kg × v2
ev
-1
2
5kgms
2kg
v = −
=
-2.5 ms-1
Here,
the velocity of the gun is negative, i.e. the gun
will
move backward.
Ans:
backward velocity=2.5 ms-1
Here,
mass
of the bullet, m1 = 10 g
=
10× 10-3 kg
=
10-2 kg
mass
of the gun, m2 = 2 kg
initial
velocity of the bullet, u1 = 0 ms-1
initial
velocity of the gun, u2 = 0 ms-1
final
velocity of the bullet, v1 = 500 ms-1
final
velocity of the gun, v2 =?
