State the principle of conservation of energy - Leaving Cert Physics - Question a - 2005

To find the maximum kinetic energy (KE) of the ball just before it hits the ground, we use the principle of energy conservation where potential energy (PE) is converted to kinetic energy (KE).

1. Calculate the potential energy at the height of 3.05 m:

   $PE = mgh$

   Where:

   • m = 0.6 kg (mass of the basketball)
   • g = 9.8 m/s² (acceleration due to gravity)
   • h = 3.05 m

   $PE = 0.6 imes 9.8 imes 3.05 = 17.94 ext{ J}$

   Therefore, the maximum kinetic energy of the ball just before it reaches the ground is:

   $KE = PE = 17.94 ext{ J}$

When the ball bounces from the ground and loses 6 joules of energy, the energy is transformed into other forms such as heat or sound. This means part of the energy is dissipated in forms that do not contribute to the ball's kinetic energy.

To calculate the height of the first bounce, we need to determine the retained energy after losing 6 joules:

$E_{retained} = E - 6 ext{ J}$ Where:

• E (max KE before the bounce) = 17.94 J
• Therefore,

$E_{retained} = 17.94 - 6 = 11.94 ext{ J}$

Now, to calculate the height using the retained energy:

$h = \frac{E}{mg}$

Where:

• m = 0.6 kg
• g = 9.8 m/s²

$h = \frac{11.94}{0.6 imes 9.8} = 2.02 ext{ m}$

Therefore, the height of the first bounce is approximately 2.02 m (acceptable range between 2.02 m to 2.03 m).