Possibility to shorten the solving procedure (q13)

[jack1234]

For this question:
http://tinyurl.com/294gl5

I am doing in this way:
[1] calculate the kinetic energy before(1/2*m1*u1 + 1/2*m2*u2)
[2] use the conservation of momentum to calculate the velocity after(v=(m1u1+m2u2)/(m1+m2))
[3] calculate the kinetic energy after(1/2*(m1+m2)*v)
this way we can get the answer, it is c

I wish to know is there a way to shorten this procedure by employing the the relationship ke=(p)^2/2m? Of course the answer can be no :)

 

[learningphysics]

Yeah, but be careful of shortcuts... be aware of what you're doing...

we can't use p^2/(2m) before the collision... p^2/(2m) is only for 1 mass (in other words we can't use p as the sum of the two momenta and m as the sum of the two masses... that won't give the right answer. but for each individual mass (1/2)mv^2 = p^2/(2m).

the initial momentum = 2*5 + 3*2 = 16kg*m/s

initial energy = (1/2)2*5^2 + (1/2)3*2^2 = 31J

we can use p^2/2m for after the collision, since after the collision we only have 1 mass of 5kg.

16^2/(2m) = 16^2/(2*5) = 25.6J

so it's a little quicker as you don't need to get the velocity...31-25.6= 5.4J

 

[ogb p]

I appreciate your curiosity and desire to find more efficient solutions. However, in this case, it is not possible to shorten the solving procedure by using the relationship ke=(p)^2/2m. This is because the conservation of momentum equation already takes into account the masses of the objects involved, so using the relationship would not provide any additional simplification. Additionally, the procedure you have described is already a simplified version of the more complex equations involved in calculating kinetic energy and momentum. So while it may not be possible to further shorten the procedure, your current approach is a valid and efficient way to solve the problem.