Final answer:
The vertical distance between two bodies of different masses tied to a string and passing over a smooth pulley after moving for one second can be calculated using Newton's Laws and kinematic equations, resulting in a distance of 0.98 meters or 98 cm.
Step-by-step explanation:
If two bodies of masses 300gm and 200gm are tied to the ends of a light string passing over a smooth pulley, and they start from the same horizontal level, we can determine the vertical distance traveled after one second using the principles of mechanics. These principles state that the only force acting upon the masses, apart from the tension in the string which is equal and opposite for both masses, is the force due to gravity (their weight).
Since the pulley is smooth and the string light, we consider the system to be frictionless. The larger mass (300gm) will accelerate downward, while the smaller mass (200gm) will accelerate upward, with the same magnitude of acceleration due to the tensions being equal (Newton's Third Law).
We can calculate the acceleration of the system by using the equation \( a = \frac{g(m_1 - m_2)}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses and \( g \) is the acceleration due to gravity. Plugging in the values, we have \( a = \frac{9.8 m/s^2 (0.3 kg - 0.2 kg)}{0.3 kg + 0.2 kg} = 1.96 m/s^2 \). Now, using the kinematic equation for motion under uniform acceleration (\( s = ut + \frac{1}{2}at^2 \)), and knowing they start from rest (\( u = 0 \)), the vertical distance (\( s \)) traveled in 1 second would be \( s = 0 m/s + \frac{1}{2}(1.96 m/s^2)(1 s^2) = 0.98 m \), or 98 cm.