A force acting on a body of mass 2kg varies with the time as shown in figure 5.11 find impulse of the force and final velocity of the body
Answers
Explanation:
For
t
=
0
to
t
=
2
, Force equation can be written in the standard form
y
=
m
x
+
c
→
F
1
(
t
)
=
t
2
Let
m
be mass of body. From Newton's second Law of motion.
→
a
(
t
)
=
1
m
×
t
2
Acceleration
→
a
≡
d
→
v
d
t
∴
→
v
(
t
)
=
1
m
∫
t
2
d
t
⇒
→
v
(
t
)
=
1
m
(
t
2
4
+
C
)
where
C
is constant of integration.
At
t
=
0
,
→
v
(
t
)
=
→
p
m
⇒
→
p
m
=
C
→
v
(
t
)
=
1
m
(
t
2
4
+
→
p
)
.....(1)
Velocity at
t
=
2
is found as
→
v
(
2
)
=
1
m
(
1
+
→
p
)
......(2)
Force equation for
t
=
2
and thereafter becomes
→
F
2
(
t
)
=
−
1
2
t
+
2
→
a
2
(
t
)
=
−
1
m
(
t
2
−
2
)
→
v
2
(
t
)
=
−
1
m
∫
(
t
2
−
2
)
d
t
→
v
2
(
t
)
=
−
1
m
(
t
2
4
−
2
t
+
C
1
)
where
C
1
is constant of integration
Using (2) to find out
C
1
→
v
2
(
2
)
=
−
1
m
(
−
3
+
C
1
)
=
1
m
(
1
+
→
p
)
⇒
(
−
3
+
C
1
)
=
−
1
−
→
p
⇒
C
1
=
2
−
→
p
∴
→
v
2
(
t
)
=
−
1
m
(
t
2
4
−
2
t
+
2
−
→
p
)
......(3)
Imposing the given condition and solving for
t
−
→
p
=
t
2
4
−
2
t
+
2
−
→
p
t
2
−
8
t
+
8
=
0
Solution of the quadratic gives us
t
=
8
±
√
64
−
4
×
1
×
8
2
t
=
4
±
2
√
2
Taking only
−
v
e
sign for original question.
For magnitude of momentum we have two solutions as above.