Solving Momentum and Kinetic Energy for 2 Blocks on Frictionless Track

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The discussion focuses on a physics problem involving two blocks on a frictionless track, where block m1 (5.00 kg) is released from a height and collides with block m2 (10.0 kg) at rest. The conservation of momentum is applied to determine their final speed after the collision, with calculations indicating a speed of approximately -3.30 m/s for the combined mass. The conservation of kinetic energy is also considered, although the energy lost during the collision remains unresolved. The conversation emphasizes the importance of understanding momentum conservation and invites further exploration of the problem, including the potential effects of friction. Overall, the discussion encourages learning through problem-solving and collaboration.
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Homework Statement


Two blocks are free to slide on a frictionless wooden track ABC shown in figure. The block of mass m1=5.00kg is released from A. The block of mass m2 = 10.0kg is initially at rest. The block of mass m1 collides with m2 and they start moving together.
physics.jpg



Homework Equations


What will be their speed after they started moving together? How much energy is lost as a result of the collision?


The Attempt at a Solution


I know we use the conservation of momentum ( m_1*v_1 = m_1*v_f + m_2*v_2). I also know that this is an elastic equation where we can use the equation for conservation of kinetic energy.
 
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so push forward, The curvature of the track simply converts the falling velocity, ie, if dropped out a window, to a horixontal one. Momentum is conserved. So things slow as you bang up. See what you can do in terms of actual numbers and eqns. No harm in being wrong, we can guide and you may learn a bunch in the process of being wrong. Personally I learn best that way
 
I think I figured out the speed after they started moving.
I used:
1/2*m_1*v_1^2 = m_1gh
v_1 = root(2*9.8*5.00)
v_1 = 9.90 m/s
v_1f = (m_1 -m_2) / (m_1 + m_2)
v_1f = -1/3 * 9.80
v_1f = -3.30 m/s

Is this right?

I'm not sure how to go about in solving the energy lost.
 
The first 3 eqns look good.The fourth depends on whether M1 bounces back., or the two masses congeal to form a body of m1+m2. The problem says they move together. so purely from conservation of mo, you can figure V2. a terrific extra credit question would be to figure out under ahat conditions M1 might recoil. Hint add friction.
 
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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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