Question

a hand grenade explodes into 3 pieces each piece flying 120 degrees relative to each other.

Piece 1 has a mass of 4 kg --

Piece 2 has a mass of 8 kg & speed of 2 m/s

Piece 3 has a mass of 8 kg & speed of 2 m/s

what is the speed of piece 1?

Answer 1

Given that after the explosion, each piece flies away at 120 degrees relative to each other.

Let's assume that the 4kg piece moves along the x-axis.

So the other 2 pieces fly away at 120 deg and -120deg(240 deg) respectively.

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Consider piece 2

$v_{2}=v_{2x}i+v_{2y}j=v_{2}cos\theta i+v_{2}sin\theta j=2m/s*cos120 i+2m/s*sin120 j$

${\color{Red} v_{2}=-1i+1.732j}$

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Consider piece 3

$v_{3}=v_{3x}i+v_{3y}j=v_{3}cos\theta i+v_{3}sin\theta j=2m/s*cos240 i+2m/s*sin240 j$

${\color{Red} v_{3}=-1i-1.732j}$

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During the explosion momentum is conserved

Initial momentum = Final momentum

$Mu=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$

$20*0m/s=4kg*v_{1}+8kg*(-1i+1.732j)+8kg*(-1i-1.732j)$

$0=4*v_{1}+8*(-1i+1.732j)+8*(-1i-1.732j)$

$4*v_{1}+(-8i+13.856j)+(-8i-13.856j)=0$

$4*v_{1}+-8i+13.856j+-8i-13.856j=0$

$4*v_{1}+(-8-8)i+(13.856-13.856)j=0$

$4*v_{1}+-16i+0j=0$

$4*v_{1}=16i+0j$

$v_{1}=\frac{16i+0j}{4}$

$v_{1}=4i+0j$

ANSWER: ${\color{Red} v_{1}=4m/s}$

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