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