Conservation of Mechanical Energy Question!?

Curt Broadhead: (Hmmm...It says that "m1 hits the ground", but then part (1) says, "...just before m2 hits the ground." So which is it? A link to "the figure" sure would have been helpful.)What the conservation of mechanical energy really says is that the sum of the PE and the KE is _constant_; that is it's the same "before" the weight drops and "after" the weight drops (and just before it hits the ground):PE_before + KE_before = PE_after + KE_afterThat's really the equation you should be using. Also, you need to include the PE and KE for both m1 and m2 in the equation.So, start with the total energy "before":PE_before = (m1)g(y1_before) + (m2)g(y2_before)KE_before = 0So, Total energy "before" is:E_before = (m1)g(y1_before) + (m2)g(y2_before)Now consider the total energy "after":PE_after = (m1)g(y1_after) + (m2)g(y2_after)KE_after = ½(m1)(v1_after)² + ½(m2)(v2_after)²But we can simplify the KE_after, because the two masses are always traveling at the same speed (si! nce they're tied together), at least until the one hits the ground. So instead of using two variables "v1_after" and "v2_after", use just a single variable "v_after" to stand for the speed of both:KE_after = ½(m1)(v_after)² + ½(m2)(v_after)² = ½(m1+m2)(v_after)²So, total energy "after" is:E_after = (m1)g(y1_after) + (m2)g(y2_after) + ½(m1+m2)(v_after)²Conservation of mechanical energy says the E_before = E_after:E_before = E_after(m1)g(y1_before) + (m2)g(y2_before) = (m1)g(y1_after) + (m2)g(y2_after) + ½(m1+m2)(v_after)²Now we use the fact that m1 fell a known distance (h=1.83 meters), which means that y1_beforeâˆ'y1_after = h. Also, m2 ROSE by that same amount, so y2_beforeâˆ'y2_after = âˆ'h. We can rewrite the above equation as:(m1)g(y1_beforeâˆ'y1_after) + (m2)g(y2_beforeâˆ'y2_after) = ½(m1+m2)(v_after)²And substitute to get:(m1)g(h) + (m2)g(âˆ'h) = ½(m1+m2)(v_after)²You can solve to that for "v_after", and there's your answer for Part 1.For Part 2, I th! ink they're saying that m2 continues flying upward for a bit (! with a slack rope), due to its momentum, after m1 hits the ground. In that case, use the conservation of energy just on m2 for this part, since m1 is now out of the picture....Show more

Darwin Ecton: m2 is 4.3 kg.