derive the Third law of motion in equation of class 9 th
Answers
Answer:
Action to an object is an opposite and equal reaction.
Answer:
Explanation:
A before and after the
collision are mA
uA
and mA
vA
, respectively. The
rate of change of its momentum (or FAB)
during the collision will be
( ) A A −
A
v u
m
t
.
Similarly, the rate of change of momentum of
ball B (= FBA) during the collision will be
( ) B B −
B
v u
m
t
.
According to the third law of motion,
the force FAB exerted by ball A on ball Band the force FBA exerted by the ball B on ball
A must be equal and opposite to each other.
Therefore,
FAB = – FBA (9.6)
or
( ) A A −
A
v u
m
t
= –
( ) B B −
B
v u
m
t
.
This gives,
mA
uA
+ mB
uB
= mA
vA
+ mB
vB
(9.7)
Since (mA
uA
+ mB
uB
) is the total momentum
of the two balls A and B before the collision
and (mA
vA
+ mB
vB
) is their total momentum
after the collision, from Eq. (9.7) 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.