In another interaction, a plasticine sphere of mass 0.30 kg is dropped 171 (c) from a height of 5.00 m and sticks to the ground. Deduce the momentum of the sphere just before impact. (i) (ii) State and explain the transfer of energy and momentum after impact.

Answer 1

(i) To deduce the momentum of the sphere just before impact, we need to use the formula:

Momentum (p) = mass (m) x velocity (v)

Given that the mass of the sphere is 0.30 kg, we need to find its velocity just before impact. We can use the equation for the velocity of an object in free fall:

v = √(2gh)

where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (5.00 m). Plugging in the values:

v = √(2 * 9.8 * 5.00) = √(98) = 9.90 m/s

Now we can calculate the momentum:

Momentum (p) = 0.30 kg * 9.90 m/s = 2.97 kg*m/s

Therefore, the momentum of the sphere just before impact is 2.97 kg*m/s.

(ii) After the impact, the energy and momentum transfer can be explained as follows:

- Energy transfer: The potential energy of the sphere before it is dropped is converted into kinetic energy as it falls. This kinetic energy is then transferred to the ground upon impact. Some of the energy is dissipated in the form of sound and heat due to the deformation of the sphere and the ground.

- Momentum transfer: Before impact, the sphere has momentum due to its velocity. Upon impact, the sphere sticks to the ground, resulting in a sudden change in momentum. The momentum of the sphere is transferred to the ground, causing the ground to experience an equal and opposite momentum change. This transfer of momentum is a result of the conservation of momentum principle, which states that the total momentum of a system remains constant unless acted upon by external forces.

Overall, the energy is transferred from potential energy to kinetic energy and then partly dissipated, while the momentum of the sphere is transferred to the ground.