I just learn about the momentum and it conservative attribute in a closed and isolate system and there is something I don't understand when I relate it to the conservation of the energy
Considering the case below:
a Block $A$ having the mass $m$ is moving at the speed $v_1$ then it hit the block $B$ having the mass $M$ with no initial speed. After the collision. Block A is sticked with block B and moving at the speed $v_2$. Assumed that there is no drag and frictional force, the block is on ground and have no gravitational potential energy. Those two block moving with the same direction and speed. According to the conservation of energy and momentum:
$mv_1 = (m+M)v_2$ (1)
$\frac12mv_1^2 = \frac12(m+M)v_2^2$ (2)
$\to mv_1^2 = (m+M)v_2^2$
because those 2 block moving with the same direction so the vector notation here is needless. After the collision there is no displacement in height of 2 block so there is no change in potential energy. Only the kinetic energy of $m$ is transform to $M$ so $M$ got speed $v_2$ and $m$'s speed decrease from $v_1$ to $v_2$
if I multiply both side of (1) to $v_1$ then
$$mv_1^2=(m+M)v_2v_1$$
However $v_1v_2$ isn't equal $v_1^2$
Can anybody explain why is it. I don't understand why
Moreover. For which cases using Momentum is advantageous. I means we can use the conservation of energy to calculate the changing of the velocity and the energy transformation of the system. So in which case it can't be used so we must use momentum to calculate.
Best Answer
Let's make a concrete example with numbers:
Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$
Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an only momentum is conserved:
After the collision velocity would be anyway lower as KE should be distributed among more mass, but some KE is lost in the crash. How much?
Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity: $$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_k = 0.5 * 2^2 *3 = 6 \rightarrow E_a = 2 + E_b = 4$.
Some energy has been transferred to B, but two thirds of the kinetic energy have been changed to other forms of energy. The general law of 'conservation of energy' has not been, anyway, violated
Velocity of center of mass is the same, although KE has changed.
Please note that momentum is conserved because we are assuming that on the surface of contact there is no friction.
A change of KE without a change of momentum is not only possible but very frequent, because p = mv momentum varies linearly and KE quadratically. You can get the same product by a wide range of factors: 6 = 6*1, = 3*2, = 2*3, = 1*6, = 0.5*12, etc., different factors give same momentum
All these factors give same values for m*v, but as the figure for v must be squared, you get all different values between momentum and energy, therefore the same factors give momentum = 6, but KE =3, =6, =9, =18, =72, etc, same momentum corresponds to many different values of KE
You have just seen that conservation of momentum is vital in inelastic collisions because energy is not conserved