Answer: Our notation is as follows : the mass of the original body is M=20.0kg ; its initial velocity is
ν
0
=(200m/s)
i
^
; the mass of one fragment is m
1
=10.0kg ; its velocity is
ν
1
=(100m/s)
j
^
; the mass of the second fragment is m
2
=4.0kg ; its velocity is
ν
2
=(−500m/s)
i
^
; and , the mass of the third fragment is m
3
=6.00kg . Conservation of linear momentum requires
M
ν
0
=m
1
ν
1
m
2
ν
2
+m
3
ν
3
.
The energy released in the explosion is equal to ΔK , the change in kinetic energy .
(a) Using the above momentum -conservation equation leads to
ν
3
=
m
3
M
ν
0
−m
1
ν
1
−m
2
ν
2
=
6.00kg
(20.0kg)(200m/s)
i
^
−(10.0kg)(100m/s)
j
^
−(4.0kg)(−500m/s)
i
^
=(1.00×10
3
m/s)
i
^
−(0.167×10
3
m/s)
j
^
.
The magnitude of
ν
3
is
ν
3
=
(1000m/s)
2
+(−167m/s)
2
=1.01×10
3
m/s
It points at θ=tan
−1
(−167/1000)=−9.48
∘
(that is at 9.5
∘
measured clockwise from the +x axis) .
(b) The energy released is ΔK :
ΔK=K
f
−K
i
=(
2
1
m
1
ν
1
2
+
2
1
m
2
ν
2
2
+
2
1
m
3
ν
3
2
)−
2
1
Mν
0
2
=3.23×10
6
J
Explanation:
