Simple but difficult for me. momentum, collsion

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Two identical bodies collide on a frictionless surface, one moving at 1 m/s and the other at 2 m/s, and they stick together after the collision. The user attempts to apply momentum conservation principles but struggles with the calculations. They initially calculate the total momentum as 3 m/s but realize that this does not yield a valid answer from the provided options. After further attempts, they suggest that the combined velocity could be 1.5 m/s but are informed that this is also incorrect. The discussion centers on clarifying the correct approach to solving the problem using momentum conservation.
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Homework Statement



Two identical bodies are sliding toward each other on a frictionless surface. One moves at 1 m/s and the other at 2 m/s. They collide and stick. The magnitude of the velocity of the combined mass is
A. 3/4 m/s.
B. 2/3 m/s.
C. 1/6 m/s.
D. 1/2 m/s.
E. 1.5 m/s.
F. none of these

Homework Equations



p1=m1(v1f-v1i)+m2(v2f-v2i)=vf(m1+m2) i think that's right. {(?))

The Attempt at a Solution



ohkay, so using the idea behind the equation i just wrote.. i did

m's are irrelevant since theyre the same

so 1m/s + 2m/s = 3m/s
so at first thought i say okay its 3/3, so 1, that's not an option, so i picked F. Wrong.

then i did m(1) +m(2) = vf (m+m)

i picked 3/2, maybe the velocity over 2 masses? so 3/2=1.5m/s

Wrong. what the heck am i doing wrong, can anyone tell me how to approach a situation like this?
 
Physics news on Phys.org
hello everyone, please excuse this, as i accidentally posted it twice =X
 
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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