Momentum Conservation => EXPLOSION

[TwinGemini14]

A projectile of mass M is moving in the +x direction with speed V when it explodes into two fragments: a lighter one having mass M/4 and a heavier one having mass 3M/4. The heavier fragment moves in the -y direction with speed V.

What is the speed of the lighter fragment? (Assume there are no external forces acting on the system).

A) V
B) 2V
C) 3V
D) 4V
E) 5V

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So the momentum after is equal to the momentum before.

So... mVbx = (1/4)mvx + (3/4)mVx
mVbx = (1/4)mvx + (0)
mVbx = (1/4)mvx => vx = 4Vb

So... mvby = (1/4)mvy + (3/4)mVy => vy = Vb

Thus... 4vb+vb = 5vb

The answer is E, correct? Can somebody please look over this and confirm whether this is right or wrong? I'm not too confident. Thanks guys!

 

[jambaugh]

The answer is correct but your work as best I can follow is not (Which is why I don't give multiple choice tests).

Write your momentum equation as a vector and then identify components. Then you will solve for x and y components of the lighter mass' velocity. Remember then that the speed is the magnitude of this vector so take the square root of the sum of the squares of the components.

 

[nlightner]

Yes, your answer is correct. According to the principle of conservation of momentum, the total momentum before the explosion (in the +x direction) must be equal to the total momentum after the explosion (taking into account the velocities and masses of the fragments in both the x and y directions). By setting up and solving the equations as you did, you correctly determined that the speed of the lighter fragment (Vb) must be equal to 4 times the speed of the original projectile (V). Therefore, the correct answer is E) 5V. Good job!