If momentum is increased by 20% then K.E. increases by _.

Explanation

The increase in momentum is given as 20%, which can be expressed as a factor of 1.2. Therefore, the new momentum is 1.2 times the original momentum. Now, the kinetic energy (K.E.) is given by the equation K.E. = (p2)/(2m), where p is the momentum and m is the mass of the object. If the momentum is increased by a factor of 1.2, then the new kinetic energy is: K.E.' = ((1.2p)2}/(2m) = (1.44p2)/(2m) The increase in kinetic energy is given by: ΔK.E. = K.E.' - K.E. ΔK.E. = (1.44p2)/(2m) - (1p2)/(2m) ΔK.E. = (0.44p2)/(2m) Since p2 and 2m are constants, we can see that the increase in kinetic energy is directly proportional to p2/(2m), which is the initial kinetic energy. Therefore, the increase in kinetic energy is: ΔK.E./K.E. = (0.44p2)/(2m) / (p2)/(2m) = 0.44 Hence, the increase in K.E. is 0.44