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Suppose there are two objects. The first object is stationary and has the next parameters: v₁=0 m/s, m₁= 4kg. The second object is heading towards the first one with speed v₂= 8m/s and mass m₂= 60 kg. Due to collision, objects merge into one and continue to move with the velocity v₃. Suppose all the energy gets converted into speed. I want to know the magnitude of velocity v₃. My question is : Why do we get slightly different results when using kinetic energy compared to conservation of momentum when solving for v₃. If we assume no energy gets converted into heat, couldn't we set up equations like : E_k1=0 J and E_k2=$frac{m_2*v_2^2}{2}$ . E₃=E₂+E₁=$frac{v_3^2}{2}*(m_1+m_2)$ . v₃=$sqrt{frac{(E_1+E_2)*2}{(m_1+m_2)}}$ ? Why do we get different result this way comparing to the result when we calculate v₃ using conservation of momentum?

Momentum is conserved if there are no external forces involved. This is the case in your setup, so the final velocity you get from conservation of momentum is right.

If you start with conservation of energy, you'll see that you get a different velocity.

Since physicists strongly believe in conservation of energy, this means that some of the mechanical energy went somewhere. This is a strong hint that you can not assume conservation of mechanical energy at will. If you assume both conservation of momentum and energy, the bodies can't stick together and move as one.

The "sticking" eats up some energy that is lost on the mechanical side.

The other answers are correct in saying that energy is not conserved, but I feel like they are lacking in truly showing why this is the case. So let's assume energy and momentum are both conserved in this process and see that we arrive at a contradiction.

So we have object 1 at rest and object 2 hits it and they stick together. Using momentum conservation:
$$m_2v_2=m_1v+m_2v$$

Using energy conservation (cancelling $1/2$ from each term):
$$m_2v_2^2=m_1v^2+m_2v^2$$

Now for fun tricks. In each of the above equations let's get all terms involving object 2 on the left and all terms involving object 1 on the right and then divide the energy equation by the momentum equation:

$$frac{m_2(v_2^2-v^2)}{m_2(v_2-v)}=frac{m_1v^2}{m_1v}$$

Now, $v_2^2-v^2=(v_2-v)(v_2+v)$, therefore (cancelling many things)

$$v_2+v=v$$
or
$$v_2=0$$

Uh oh. We see that before the collision object 2 is at rest. But object 1 was at rest also$^*$. Therefore no collision with these properties could have occurred. Therefore something is wrong with our assumptions of both energy and momentum conservation.

Momentum has to be conserved since there are no external forces on the system. Therefore it must be that energy is not conserved. There is nothing saying energy must be conserved here, but we have now shown that it actually cannot be conserved.

$^*$ In general if we had specified the velocity of $m_1$ by $v_1$ we would have arrived at $v_1=v_2$ using similar methods used above. This just means the objects started out with the same velocity and never actually collide. Therefore, the only way our system can conserve both energy and momentum and have equal "final" velocities is if no collision happened.

The objects are merged, and an inelastic collision is one in which objects stick together after impact: kinetic energy is not conserved. Therefore, you can't assume that kinetic energy is conserved; merging takes a little bit of energy.

If this system is isolated, and no force acted upon neither of the two objects, then, momentum is conserved; hence, you can calculate the final velocity using the momentum's equation.

Suppose all the energy gets converted into speed.

Uh oh! That's your error. You can't just suppose that something that doesn't happen, happens.

The collision was inelastic. That means some of the energy gets converted to heat, as the objects merge. This is the difference between your total before and after energy.

I think that the hypothesis of momentum conservation is flawed since the problem really doesn't state that. Also there's nothing saying that the system is isolated at all.

Hence you calculated what you can (assuming that "energy" in the problem statement refers only to the kinetic energy itself).
If the hypothesis of kinetic energy conversion is correct (I would assume so since it's stated) then your energy-based calculation is correct :) and the missing momentum is due to an external source.