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A 37.0-$kg$ body is moving in the direction of the positive x axis with a speed of 342 $m/s$ when, owing to an internal explosion, it breaks into three pieces. One part, whose mass is 5.0 $kg$, moves away from the point of explosion with a speed of 311 $m/s$ along the positive y axis. A second fragment, whose mass is 4.0 $kg$, moves away from the point of explosion with a speed of 402 $m/s$ along the negative x axis. What is the speed of the third fragment? Ignore effects due to gravity.

Respuesta :

Answer: [tex]v_{3}[/tex] = 512.3 m/s

Explanation:

Given:

M = 37 kg

v = 342 m/s

[tex]m_{1}[/tex] = 5 kg

[tex]v_{1}[/tex] = 311 m/s

[tex]m_{2}[/tex] = 4 kg

[tex]v_{2}[/tex] = 402 m/s

∴ [tex]m_{3}[/tex] = M - [tex]m_{1}[/tex] - [tex]m_{2}[/tex]

[tex]m_{3}[/tex] = 28 kg

Now, to compute the speed of third fragment we will apply conservation of momentum:

M[tex]\times[/tex]v = [tex]m_{1}\times v_{1}[/tex] + [tex]m_{2}\times v_{2}[/tex] +[tex]m_{3}\times v_{3}[/tex]

Putting values in the equation given above:

37[tex]\times[/tex](342)[tex]\hat{i}[/tex] = 5[tex]\times[/tex](311)[tex]\hat{j}[/tex] + 4[tex]\times[/tex](402)[tex]\hat{-i}[/tex] + 28[tex]\times v_{3}[/tex]

[tex]v_{3}[/tex] = 509.3 m/s [tex]\hat{i}[/tex] - 55.53 m/s [tex]\hat{j}[/tex]

Now, the magnitude of velocity of the third particle is;

[tex]v_{3}[/tex] = [tex]\sqrt{509.3^{2} + 55.53^{2} }[/tex]

[tex]v_{3}[/tex] = 512.3 m/s

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