How Is the Initial Momentum of a 2kg Block Determined After Explosion?

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The initial momentum of the 2 kg block can be calculated using the conservation of momentum principle. The calculated momentum of 13 kg m/s is correct, as it considers the vertical and horizontal components of the exploded blocks' velocities. The 1.2 kg block moving east at 10 m/s contributes to the horizontal momentum, while the 0.8 kg block moving at 22 m/s at an angle north of west has both vertical and horizontal components that must be resolved. The misunderstanding arose from incorrectly assuming the horizontal components cancel each other out. Thus, the initial momentum of the block is confirmed to be 13 kg m/s.
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Homework Statement



A 2 kg block moves north at an unknown velocity. It then explodes into two blocks, the first one being 1.2 kg and moves EAST at 10 m/s, the second one is 0.8 kg and moves 22 m/s at an angle North of West.

What is the initial MOMENTUM of the 2kg block?

Homework Equations



p initial = p final (momentum conserved)

The Attempt at a Solution



The answer i got was 13 kg m/s, and some friends of mine got 29.6 kg m/s. The reason i got 13 is because I think/thought the two horizontal components of the exploded blocks cancel out each other because they are in separate directions (velocity is a vector and therefore one being negative one being positive, so I thought only the vertical component of the second block's momentum mattered).

Any help is appreciated Thank You!
 
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You are correct; your friend is not. Good reasoning.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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