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$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$ can change to make the kinetic energy $E_\text{kin}$ decrease while still preserving the total momentum $\vec{P}$.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$ With Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$) you further get $$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$ and hence $$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$ can change to make the total kinetic energy $E_\text{kin}$ decrease while still preserving the total momentum $\vec{P}$.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$ By applying Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$) to these two forces you further get $$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$ and hence $$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$P=m_1v_1+m_2v_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$K.E.=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$F_{2\to 1}=-F_{1\to 2}$$ With Newton's second law ($F=m\frac{\Delta v}{\Delta t}$) you further get $$m_1\frac{\Delta v_1}{\Delta t}=-m_2\frac{\Delta v_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1v_1+m_2v_2)}{\Delta t}=0$$ and hence $$m_1v_1+m_2v_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$\vec{P}=m_1\vec{v}_1+m_2\vec{v}_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$E_\text{kin}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

So there are many possible ways how the velocities $\vec{v}_1$ and $\vec{v}_2$ can change to make the kinetic energy $E_\text{kin}$ decrease while still preserving the total momentum $\vec{P}$.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$\vec{F}_{2\to 1}=-\vec{F}_{1\to 2}$$ With Newton's second law ($\vec{F}=m\frac{\Delta \vec{v}}{\Delta t}$) you further get $$m_1\frac{\Delta \vec{v}_1}{\Delta t}=-m_2\frac{\Delta \vec{v}_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1\vec{v}_1+m_2\vec{v}_2)}{\Delta t}=\vec{0}$$ and hence $$m_1\vec{v}_1+m_2\vec{v}_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$P=m_1v_1+m_2v_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$K.E.=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$F_{2\to 1}=-F_{1\to 2}$$ With Newton's second law ($F=m\frac{\Delta v}{\Delta t}$) you further get $$m_1\frac{\Delta v_1}{\Delta t}=-m_2\frac{\Delta v_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1v_1+m_2v_2)}{\Delta t}$$ and hence $$m_1v_1+m_2v_2=\text{const}$$ which is just the above mentioned conservation of total momentum.

$K.E=\frac{1}{2}mv^2$, $P=mv$ thus there is a relation between them. I really cant understand why there is no decrease in momentum when kinetic energy is decreased in inelastic collision.

It doesn't make sense to consider a single body during a collision (i.e. during interaction with a second body). You need to look at both bodies together. Therefore you need to consider the total momentum of both bodies. $$P=m_1v_1+m_2v_2 \tag{1}$$ It is this quantity which is conserved during the collision.

Likewise you need to consider the total kinetic energy of both bodies. $$K.E.=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ It is this quantity which is conserved during an elastic collision.

Should i leave this question until i learn "Lagrangian" and "Noether's Theorem" ?

To understand momentum conservation (1) you don't need Lagrangian mechanics or Noether's theorem. Newton's mechanics is just enough.

According to Newton's third law (actio = reactio) you have $$F_{2\to 1}=-F_{1\to 2}$$ With Newton's second law ($F=m\frac{\Delta v}{\Delta t}$) you further get $$m_1\frac{\Delta v_1}{\Delta t}=-m_2\frac{\Delta v_2}{\Delta t}$$

Rearranging this you get $$\frac{\Delta(m_1v_1+m_2v_2)}{\Delta t}=0$$ and hence $$m_1v_1+m_2v_2=\text{const}$$ which is just the above mentioned conservation of total momentum.