When a rocket takes off the total momentum of the system is zero but the total energy is not zero explain why?
Answers
Explanation:
is
→
p
f
=
→
p
rocket
+
→
p
gas
=
(
m
−
d
m
g
)
(
v
+
d
v
)
^
i
+
d
m
g
(
v
−
u
)
^
i
.
Since all vectors are in the x-direction, we drop the vector notation. Applying conservation of momentum, we obtain
p
i
=
p
f
m
v
=
(
m
−
d
m
g
)
(
v
+
d
v
)
+
d
m
g
(
v
−
u
)
m
v
=
m
v
+
m
d
v
−
d
m
g
v
−
d
m
g
d
v
+
d
m
g
v
−
d
m
g
u
m
d
v
=
d
m
g
d
v
+
d
m
g
v
.
Now,
d
m
g
and dv are each very small; thus, their product
d
m
g
d
v
is very, very small, much smaller than the other two terms in this expression. We neglect this term, therefore, and obtain:
m
d
v
=
d
m
g
u
.
Our next step is to remember that, since
d
m
g
represents an increase in the mass of ejected gases, it must also represent a decrease of mass of the rocket:
d
m
g
=
−
d
m
.
Replacing this, we have
m
d
v
=
−
d
m
u
or
d
v
=
−
u
d
m
m
.
Integrating from the initial mass mi to the final mass m of the rocket gives us the result we are after:
∫
v
v
i
d
v
=
−
u
∫
m
m
i
1
m
d
m
v
−
v
i
=
u
ln
(
m
i
m
)
and thus our final answer is
Δ
v
=
u
ln
(
m
i
m
)
.