To calculate the change in momentum of the ball, we need to consider the initial momentum and the final momentum. The momentum of an object is given by the product of its mass and velocity.
Given: Mass of the ball, m = 0.5 kg Initial height, h_i = 3.2 m Final height, h_f = 1.8 m
First, let's calculate the initial velocity (v_i) of the ball when it was dropped from rest at a height of 3.2 m. We can use the equation for gravitational potential energy:
Potential energy (PE) = mgh
Where: m = mass of the ball g = acceleration due to gravity (approximately 9.8 m/s^2) h = height
PE_i = m * g * h_i PE_i = 0.5 kg * 9.8 m/s^2 * 3.2 m PE_i = 15.68 J
Since the ball was dropped from rest, all of the initial potential energy is converted to kinetic energy:
KE_i = PE_i
Using the equation for kinetic energy:
KE_i = (1/2) * m * v_i^2
15.68 J = (1/2) * 0.5 kg * v_i^2 31.36 J = 0.25 kg * v_i^2 v_i^2 = 31.36 J / 0.25 kg v_i^2 = 125.44 m^2/s^2 v_i ≈ 11.2 m/s (taking the positive square root)
Now, let's calculate the final velocity (v_f) of the ball when it reached a height of 1.8 m. Since the ball bounces back, we assume that the final velocity has the same magnitude as the initial velocity, but in the opposite direction:
v_f = -v_i v_f = -11.2 m/s
The change in momentum (Δp) is given by the difference between the final momentum (p_f) and the initial momentum (p_i):
Δp = p_f - p_i
p_i = m * v_i p_i = 0.5 kg * 11.2 m/s p_i = 5.6 kg·m/s (upwards)
p_f = m * v_f p_f = 0.5 kg * (-11.2 m/s) p_f = -5.6 kg·m/s (downwards)
Δp = (-5.6 kg·m/s) - (5.6 kg·m/s) Δp = -11.2 kg·m/s
Therefore, the change in momentum of the ball is -11.2 kg·m/s. It's important to note that the negative sign indicates a change in direction, as the ball bounces back upwards. So, the answer 7 kg·m/s provided by your textbook seems incorrect. The correct answer is -11.2 kg·m/s.