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