Energy

Relation between kinetic energy and momentum
 
Let us consider a body of mass ‘m’ having a velocity ‘v’ , then 
momentum of the body     P = mass × velocity  \(\rightarrow\)P = m × v
                              \( \Rightarrow v = \frac{P} {m} \)      --------- (1)
From definition, kinetic energy (K.E) of the body 
                    K.E =  \( K.E = \frac{1} {2}mv^2 \)           --------- (2)
Now putting the value of (1) in (2) we have 
               \( K.E = \frac{1} {2}m\left( {\frac{P} {m}} \right)^2 \)     
                    K.E. = \( K.E = \frac{1} {2}m\frac{{P^2 }} {{m^2 }} = \frac{1} {2}\frac{{P^2 }} {m} = \frac{{P^2 }} {{2m}} \)  ----- (3)
Thus we can write P² = 2m × K.E  \( \Rightarrow P = \sqrt {2m \times K.E} \)  
Thus momentum = \( \sqrt {2 \times mass \times kinetic energy} \)
 
Note :
1. For same momentum \( K.E\alpha \frac{1} {m} \)  Kinetic energy varies inversely as the mass.
2. If two bodies have same momentum, ratio of their kinetic energy is 
             \( \frac{{K.E_1 }} {{K.E_2 }} = \frac{{m_2 }} {{m_1 }} \)     \( \left[ {E\alpha \frac{1} {m}} \right] \)          
3. If two bodies have same kinetic energy, ratio of their momenta is \( \frac{{P_1 }} {{P_2 }} = \sqrt {\frac{{m_1 }} {{m_2 }}} \)
4. If the momentum of a body is increased to x times, its kinetic energy increases to x² times. 
5. If the kinetic energy of a body is increased to x times, its momentum increases to \(\sqrt x\) times.