A 22 kg body is moving through space in the positive direction of an x axis with a speed of 190 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 13 kg, moves away from the point of explosion with a speed of 130 m/s in the positive y direction. A second part, with a mass of 2.2 kg, moves in the negative x direction with a speed of 460 m/s. What are the (a) x-component and (b) y-component of the velocity of the third part

Answers

Answer 1

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:


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[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Let's consider the given terms ~

Initial velocity (u) = 100 m/s

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[tex]v = u + at[/tex]

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[tex]t = 10[/tex]

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Now, let's find how much distance (s) it will cover before it stops using second equation of motion ~

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Answers

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