Solve Collision Problem: Two 8kg & 5.9kg Blocks

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Two blocks with masses of 8 kg and 5.9 kg are on a frictionless surface, separated by a spring. When a cord holding them together is burned, the 5.9 kg block moves to the right at 20 m/s. The conservation of momentum equation is applied to find the velocity of the 8 kg block. Initially at rest, the final velocity of the 8 kg block is calculated to be 14.75 m/s, but the direction of motion must be considered, resulting in a negative value. The correct final velocity of the 8 kg block is determined to be -14.75 m/s, indicating it moves in the opposite direction.
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1. Two blocks of masses 8 kg and 5.9 kg are
placed on a horizontal, frictionless surface. A
light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. A cord holding them together
is burned, after which the block of mass 5.9 kg
moves to the right with a speed of 20 m/s.
What is the velocity of the other mass in
m/s?
Answer in units of m/s.



2. m1v1f+m2v2f=m1v1i+m2v2i



3. the blocks of masses m sub 1 and m sub 2 are connected by a cord with a spring between them. i assumed that they are initially at rest due to the cord. so the initial portion of the above equation would be 0 as their velocities would be 0. solving for v sub 2 final i get m1v1f/m2 =v2f. i get 14.75 m/s. this is incorrect.
 
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Maybe you're failing to take the direction of the velocity into account.
 
yup that was it lol. moving in opposite direction give opposite signs
 
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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