Question

Two masses are connected by a light string passing over a light, frictionless pulley, as shown in the figure below. The object of mass m1 is released from rest at height h above the table. Use the isolated system model to answer the following. (We assume m1 > m2.)

(a) Determine the speed of m2 just as m1 hits the ground. (Use any variable or symbol stated above along with the following as necessary: g.)

v =

(b) Find the maximum height above the ground to which m2 rises. (Use any variable or symbol stated above along with the following as necessary: g.)

hmax =

Answer 1

(a)
Initial energy of the system
= potential energy of the mass 1 = m₁gh ---------(1)
Now final energy of the system when mass 1 hits the floor
Final energy
= KInetic energy of mass 1 and 2 + potential energy of mass 2
= (1/2)(m₁ +m₂)V² + m₂gh -------------(2)
Applying the energy conservation
Initial = final energy
(1/2)(m₁ +m₂)V² + m₂gh = m₁gh
(1/2)(m₁ +m₂)V² = (m₁ - m₂)gh


(b) Now the kinetic energy of the mass 2 when mass 1 hits the floor is
K = (1/2)mV²
Now this energy will consumed by increasing the potential energy , therefore
At a height say h₂ it will get consumed, therefore
(1/2)m₂V² = m₂gh₂
h₂ = (1/2g)V²
Hence the maximum height from the ground
h_max = h + h₂