A light string passes over a smooth pulley - HSC - SSCE Mathematics Extension 2 - Question 6 - 2024 - Paper 1

To find the acceleration of the 9 kg mass, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Assuming downward is positive and using the gravitational force on the masses, we can write the equations of motion for both masses:

1. For the 9 kg mass:

   $F_{net, 9kg} = m_{9kg} g - T = m_{9kg} a$

   Where:

   • $m_{9kg} = 9$ kg (mass of the first object)
   • $T$ = tension in the string
   • $a$ = acceleration of the 9 kg mass

2. For the 5 kg mass:

   $T - m_{5kg} g = -m_{5kg} a$

   Where:

   • $m_{5kg} = 5$ kg (mass of the second object)

Setting the tensions equal, we can solve the equations:

Adding both equations to eliminate $T$:

$m_{9kg} g - m_{5kg} g = (m_{9kg} + m_{5kg}) a$

Substituting values:

$9g - 5g = (9 + 5) a$

This simplifies to:

$4g = 14a$

Now, solve for $a$:

a = rac{4g}{14} = rac{2g}{7}

Thus, the acceleration of the 9 kg mass is $\frac{2g}{7}$.