A ball of mass 100g fall from a height of 3m onto a horizontal surface and rebounds to a height of 2m. Calculate:(a) in momentum and (b) in kinetic energy.
K.E. = m g h
momentum = m √(2 g h)
To calculate the momentum and kinetic energy, we can use the following equations:
(a) Momentum (p) = mass (m) × velocity (v)
(b) Kinetic Energy (K.E.) = 0.5 × mass (m) × velocity squared (v²)
First, let's find the velocity of the ball before and after the rebound. We can use the law of conservation of energy. The potential energy (PE) before the fall is equal to the sum of the kinetic energy (K.E.) and potential energy (PE) after the rebound:
PE before = K.E. + PE after
The potential energy before the fall is given by the formula:
PE before = m × g × h
where m is the mass, g is the acceleration due to gravity (9.8 m/s²), and h is the height.
In this case, m = 100g = 0.1 kg, h = 3m, and g = 9.8 m/s².
So, PE before = 0.1 kg × 9.8 m/s² × 3m = 0.294 J
Next, let's calculate the potential energy after the rebound:
PE after = m × g × h
where h is the height to which the ball rebounds.
In this case, h = 2m.
So, PE after = 0.1 kg × 9.8 m/s² × 2m = 0.196 J
Using the conservation of energy equation, we can now find the initial velocity (v) of the ball before the fall:
PE before = K.E. + PE after
0.294 J = K.E. + 0.196 J
K.E. = 0.294 J - 0.196 J = 0.098 J
Now, let's calculate the velocity (v) using kinetic energy:
K.E. = 0.5 × m × v²
0.098 J = 0.5 × 0.1 kg × v²
v² = (0.098 J) / (0.5 × 0.1 kg) = 1.96 m²/s²
v = √(1.96 m²/s²) = 1.4 m/s (rounded to one decimal place)
Now, we can calculate the momentum (p) using the initial velocity:
p = m × v = 0.1 kg × 1.4 m/s = 0.14 kg·m/s
Therefore, the momentum of the ball is 0.14 kg·m/s.
Finally, we can calculate the kinetic energy (K.E.) using the final velocity:
K.E. = 0.5 × m × v²
K.E. = 0.5 × 0.1 kg × (1.4 m/s)² = 0.098 J
Therefore, the kinetic energy of the ball is 0.098 Joules.