Question

Two masses are connected by a light string passing over a light, frictionless pulley as in Figure P5.63. The m1 = 5.30 kg object is released from rest at a point 4.00 m above the floor, where the m2 = 3.35 kg object rests.

Answer 1

I think question is incomplete. so rest question

(a) Determine the speed of each object when the two pass each other.
(b) Determine the speed of each object at the moment the 5.30 kg mass hits the floor.
(c) How much higher does the 3.35 kg mass travel after the 5.30 kg mass hits the floor?

Apply conservation of energy to write an equation for each block separately, add them together, and solve for final velocity, v. For m₁, it initially has some gravitational PE of m₁gh and zero KE. When it arrives at a point even with m₂, it has fallen half its original height of 4.00m, or 2.00m. At that time half it initial PE is now KE, therefore:

m₁gh = 0.5m₁v² + m₁gh / 2
2m₁gh = m₁v² + m₁gh-------------->(1)

The same can be said of m₂, but it will have zero initial KE and PE, so that when it arrives even with m₁ it will have both KE (of 0.5m₂v²) and PE (of m₂gh / 2):

0 = 0.5m₂v² + m₂gh / 2
0 = m₂v² + m₂gh-------------------->(2)

Adding them together and solving for v:

2m₁gh = (m₁v² + m₁gh) + (m₂v² + m₂gh)
2m₁gh = m₁v² + m₂v² + m₁gh + m₂gh
v²(m₁ + m₂) = m₁gh - m₂gh
v² = gh(m₁ - m₂) / (m₁ + m₂)
v = sqrt[ gh(m₁ - m₂) / (m₁ + m₂)]
= sqrt[9.80m/s²(4.00m)(5.30kg - 3.35kg) / (5.30kg + 3.35kg)]
= 2.97m/s

(b) In (a), we considered a point half way up (2.00m), in this part h = 4.00m, so the equation we obtained in (a) would differ only in that it will have a coefficient of 2 on gh, so:

v = sqrt[ 2gh(m₁ - m₂) / (m₁ + m₂)]
= sqrt[2*9.80m/s²(4.00m)(5.30kg - 3.35kg) / (5.30kg + 3.35kg)]
= 4.21m/s

(c) Because m₂ is also moving at 4.21m/s when m₁ hits the floor, it will have KE and assume zero PE at this point (we’re calling this point the zero point, and it will rise some point h above that point, so:

mgh = 0.5mv²
h = v² / 2g
= (4.21m/s)² / 2(9.80m/s²)
= 0.904m