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Two masses, 2m kg and 4m kg, are attached by a light string - HSC - SSCE Mathematics Extension 2 - Question 16 - 2020 - Paper 1

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Two masses, 2m kg and 4m kg, are attached by a light string. The string is placed over a smooth pulley as shown. The two masses are at rest before being released an... show full transcript

Worked Solution & Example Answer:Two masses, 2m kg and 4m kg, are attached by a light string - HSC - SSCE Mathematics Extension 2 - Question 16 - 2020 - Paper 1

Step 1

Show that \( \frac{dv}{dt} = \frac{gm - kv}{3m} \)

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Answer

Let ( g ) be the acceleration due to gravity acting on the masses.

For the 4m mass, the forces acting on it are:

  • Weight: ( 4mg )
  • Tension: ( T )
  • Air resistance: ( kv_{4m} )

Using Newton's second law, we have:

[ 4mg - T - kv_{4m} = 4m \cdot a ]
where ( a ) is the acceleration of the system.

For the 2m mass:

  • Weight: ( 2mg )
  • Tension: ( T )
  • Air resistance: ( kv_{2m} )

Thus:

[ T - 2mg - kv_{2m} = 2m \cdot (-a) ]
(because the acceleration acts in the opposite direction)

Setting the equations for acceleration equal based on the relationship ( a = \frac{dv}{dt} ), we can derive:

  1. From the 4m mass:

[ a = \frac{4mg - T - kv_{4m}}{4m} ]

  1. From the 2m mass:

[ a = \frac{T - 2mg - kv_{2m}}{-2m} ]

By solving these equations simultaneously, we arrive at:

[ \frac{dv}{dt} = \frac{gm - kv}{3m} ]

Step 2

Show that when \( t = \frac{3m}{k} \ln 2 \), the velocity of the larger mass is \( \frac{gm}{2k} \)

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Answer

Starting with the differential equation derived earlier:

[ \frac{dv}{dt} = \frac{gm - kv}{3m} ]

Rearranging gives:

[ \frac{dv}{gm - kv} = \frac{1}{3m} dt ]

Integrating both sides, we have:

[ \int \frac{1}{gm - kv} dv = \int \frac{1}{3m} dt ]

This yields:

[ -\frac{1}{k} \ln |gm - kv| = \frac{t}{3m} + C ]

Solving for ( v ) requires determining the constant ( C ) when the mass is initially at rest (( v = 0 ) at ( t = 0 )).

Substituting values, we solve for ( v ) at the specified time:

When ( t = \frac{3m}{k} \ln 2 ):

[ v = \frac{gm}{2k} ]

Thus, we have shown the required relationship.

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