A basketball of mass 0.602 kg is passed to a player of mass 83.0 kg who jumps off the floor while catching the ball. The player's center of mass is instantaneously at rest just before he catches the ball, which is initially moving at a velocity of 11.5 m/s, directed 30.0° below the horizontal. Find the horizontal component of the center of mass velocity of the system of player and ball just after the catch.
To find the horizontal component of the center of mass velocity of the system of player and ball just after the catch, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on it. In this case, we can assume that no external forces are acting on the system of the player and the ball after the catch.
The total momentum of the system before the catch can be calculated by adding the individual momenta of the ball and the player.
Momentum of the ball (p1) = mass of the ball (m1) * velocity of the ball (v1)
p1 = 0.602 kg * 11.5 m/s * cos(30°) (since we are interested in the horizontal component)
Momentum of the player (p2) = mass of the player (m2) * velocity of the player (v2)
p2 = 83.0 kg * 0 m/s (since the player is at rest before catching the ball)
The total momentum before the catch (p_initial) is then the sum of p1 and p2.
p_initial = p1 + p2
Now, since momentum is a vector quantity, the final momentum of the system (p_final) will also be a vector sum. The magnitude of p_final will be equal to the magnitude of p_initial:
|p_final| = |p_initial|
The direction of p_final will be different from p_initial, as the player will impart a change in the ball's momentum when catching it. However, the question asks for the horizontal component of the center of mass velocity, so we are only concerned with the horizontal direction, which is unchanged by the interaction.
Therefore, the horizontal component of the center of mass velocity of the system just after the catch is the same as the horizontal component of the center of mass velocity just before the catch, which is given by the horizontal component of the ball's velocity:
horizontal component of the center of mass velocity = velocity of the ball (v1) * cos(30°)
Now, we can plug in the values and calculate the result:
v1 = 11.5 m/s
cos(30°) = √3/2
horizontal component of the center of mass velocity = 11.5 m/s * √3/2 = 5.291 m/s
Therefore, the horizontal component of the center of mass velocity of the system just after the catch is approximately 5.291 m/s.
To find the horizontal component of the center of mass velocity of the system just after the catch, we can use the principle of conservation of momentum.
Momentum is defined as the product of an object's mass and velocity, and it is conserved in a closed system where no external forces are acting.
Given:
- Mass of the basketball (m1) = 0.602 kg
- Mass of the player (m2) = 83.0 kg
- Initial velocity of the basketball (v1) = 11.5 m/s
- Angle below the horizontal (θ) = 30.0°
First, we need to calculate the horizontal (x) and vertical (y) components of the initial velocity of the ball (v1x and v1y).
v1x = v1 * cos(θ)
v1y = v1 * sin(θ)
v1x = 11.5 m/s * cos(30.0°) ≈ 9.98 m/s
v1y = 11.5 m/s * sin(30.0°) ≈ 5.75 m/s
Next, we calculate the initial momentum of the system before the catch.
Initial momentum (p1) = (m1 * v1x) + (m2 * 0)
Initial momentum (p1) = (0.602 kg * 9.98 m/s) + (83.0 kg * 0)
Initial momentum (p1) ≈ 5.99 kg*m/s
Since the system is closed and there are no external forces acting, the total momentum before the catch is equal to the total momentum after the catch.
Final momentum (p2) = (m1 * v2x) + (m2 * v2x)
The center of mass velocity of the system just after the catch depends on the velocity of both the player and the ball. However, we are only interested in the horizontal component of the center of mass velocity. This means that the player's vertical velocity will not affect our calculation.
To find the v2x, we need to set up a conservation of momentum equation.
p1 = p2
Do not include the y-component of the momentum since it does not affect the horizontal component of the center of mass velocity.
m1 * v1x = (m1 + m2) * v2x
(0.602 kg * 9.98 m/s) = (0.602 kg + 83.0 kg) * v2x
59.9996 kg*m/s = 83.602 kg * v2x
v2x = (59.9996 kg*m/s) / (83.602 kg)
v2x ≈ 0.717 m/s
Therefore, the horizontal component of the center of mass velocity of the system just after the catch is approximately 0.717 m/s.