Questions: To make a bounce pass, a player throws a 0.50-kg basketball toward the floor. The ball hits the floor with a speed of 5.0 m/s at an angle of 65° to the vertical. If the ball M=0.5 vi=5 θ=65° rebounds with the same speed and angle, what was the impulse delivered to it by the floor? (Hint: Consider x and y components of the velocity before and after the impact with the floor)

To make a bounce pass, a player throws a 0.50-kg basketball toward the floor. The ball hits the floor with a speed of 5.0 m/s at an angle of 65° to the vertical. If the ball
M=0.5
vi=5
θ=65°
rebounds with the same speed and angle, what was the impulse delivered to it by the floor? (Hint: Consider x and y components of the velocity before and after the impact with the floor)

Transcript text: 2 3. To make a bounce pass, a player throws a $0.50-\mathrm{kg}$ basketball toward the floor. The ball hits the floor with a speed of $5.0 \mathrm{~m} / \mathrm{s}$ at an angle of $65^{\circ}$ to the vertical. If the ball \[ \begin{array}{l} M=0.5 \\ v_{i}=5 \\ \theta=65^{\circ} \end{array} \] rebounds with the same speed and angle, what was the impulse delivered to it by the floor? (Hint: Consider $x$ and $y$ components of the velocity before and after the impact with the floor)

Solution

Solution Steps

Step 1: Determine the Initial and Final Velocity Components

The initial velocity of the basketball is given as $ v_i = 5.0 \, \text{m/s} $ at an angle of $ \theta = 65^\circ $ to the vertical. We need to find the components of this velocity in the $ x $ and $ y $ directions.

Initial $ x $-component of velocity: \[ v_{ix} = v_i \sin(\theta) = 5.0 \sin(65^\circ) \]
Initial $ y $-component of velocity: \[ v_{iy} = v_i \cos(\theta) = 5.0 \cos(65^\circ) \]

Since the ball rebounds with the same speed and angle, the final velocity components are:

Final $ x $-component of velocity: \[ v_{fx} = v_i \sin(\theta) = 5.0 \sin(65^\circ) \]
Final $ y $-component of velocity: \[ v_{fy} = -v_i \cos(\theta) = -5.0 \cos(65^\circ) \]

The negative sign in $ v_{fy} $ indicates that the direction of the $ y $-component of velocity is reversed after the bounce.

Step 2: Calculate the Change in Velocity Components

The change in velocity components ($\Delta v_x$ and $\Delta v_y$) are calculated as follows:

Change in $ x $-component of velocity: \[ \Delta v_x = v_{fx} - v_{ix} = 5.0 \sin(65^\circ) - 5.0 \sin(65^\circ) = 0 \]
Change in $ y $-component of velocity: \[ \Delta v_y = v_{fy} - v_{iy} = -5.0 \cos(65^\circ) - 5.0 \cos(65^\circ) = -2 \times 5.0 \cos(65^\circ) \]

Step 3: Calculate the Impulse Delivered by the Floor

Impulse is given by the change in momentum, which is the product of mass and change in velocity. The impulse has both $ x $ and $ y $ components:

Impulse in the $ x $-direction: \[ J_x = m \Delta v_x = 0.5 \times 0 = 0 \]
Impulse in the $ y $-direction: \[ J_y = m \Delta v_y = 0.5 \times (-2 \times 5.0 \cos(65^\circ)) \]

\[ J_y = -1.0 \times 5.0 \cos(65^\circ) \]

\[ J_y = -5.0 \cos(65^\circ) \]