What was wrong with my calculation?

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The discussion centers on a physics problem involving a collision between two objects, where the user correctly calculated the velocity of the 10kg object post-collision but is confused about the percentage of kinetic energy lost. The user consistently arrives at a KE loss of 7.4% instead of the expected 7.2%. They express uncertainty about the accuracy of the textbook's equation and seek clarification on the symbols used, particularly the meaning of "**" as squared. The user is frustrated and considers informing their professor about the discrepancy in the answers. The conversation highlights the importance of understanding equations and verifying calculations in physics problems.
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Homework Statement



a 8kg object moving east at 15m/s at a frictionless horizontal surface collides with 10kg object that is initially at rest. After collision, the 8kg object moves south at 4m/s.
a)what is the velocity of 10kg object after collision?
b)what percentage of the initial kinetic energy is lost in the collision?
The answer is a)12.4 m/s at 14.9 angle N of E b)KE lost=7.2%

Homework Equations



%KE lost = (KE1 - KE2)/(KE1)*100
KE1 = 1/2*m1*v1**2
KE2 = 1/2*m1*v2**2 + 1/2*m2*v3**2
Someone gave me this equation. But I cannot understand lots of symbols such as ** here.
Note:v3 is velocity of 10kg object after collision, which is 12.4m/s.


The Attempt at a Solution



I already solved a) and I was solving b) now. I pugged my value in the equation above assuming that ** means squared. And my answer was 7.4!
how could this can happen? I tried again and again and 7.4 kept came out instead of 7.2
Does my textbook is wrong of that equation is wrong? If the equation is wrong, pleas tell my the fixed form.
 
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Yes, I make it 7.4 or 7.5 also.
 
whew! I thought I made a huge mistake in the middle! I must tell my prof that the answer 7.2 is wrong.
 
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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