The second collision I did was an Inelastic collision 7. The second collision is
ID: 1265331 • Letter: T
Question
The second collision I did was an Inelastic collision 7. The second collision is an example of a perfectly inelastic collision. Such a collision involves the greatest possible loss of kinetic energy. Using conservation of momentum, show that the expected Percent Change in KF reduces to [-M2/(M1 + M2]*100 (i.e. derive this formula). Use this formula to calculate the expected Percent Change in KE. Find the percent error between this value and what you obtained previously in pert 3 of the Calculations section above. Are the two remit consistent? Discuss.
Explanation / Answer
In perfectly inelastic collision, the partciles stick together after collision, hence, having the same velocity.
Thus,
m1 v1i = (m1 + m2)v
--> v = m1 v1i / (m1 + m2)
Thus, the total KE after the collision is
KEf = 1/2 (m1 + m2) v^2
= m1^2 v1i^2 / [2(m1 + m2)]
Also, the initial KE is
KEi = 1/2 m1 v1i^2
THus,
%change KE = (KEf - KEi)/KEi
Plugging in our expressions and simplifying,
= -m2 / (m1 + m2) [ANSWER]