A 70.5kg football player is gliding across very smooth ice at 2.05m/s . He throws a 0.470kg football straight forward.
What is the player's speed afterward if the ball is thrown at 14.5m/s relative to the ground?
What is the player's speed afterward if the ball is thrown at 14.5m/s relative to the player?
14
22
To solve for the player's speed afterward in both scenarios, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided no external forces are acting on the system.
Let's first calculate the initial momentum of the football player before the ball is thrown.
Initial momentum of the football player = mass of the football player * velocity of the football player
Given:
Mass of the football player = 70.5 kg
Velocity of the football player = 2.05 m/s
Initial momentum of the football player = (70.5 kg) * (2.05 m/s)
Initial momentum of the football player = 144.025 kg·m/s
Now, let's consider the first scenario where the ball is thrown at 14.5 m/s relative to the ground.
To calculate the final momentum of the system after the ball is thrown, we can use the conservation of momentum principle.
Final momentum of the system = Initial momentum of the football player + Momentum of the thrown ball
Given:
Mass of the thrown ball = 0.470 kg
Velocity of the thrown ball relative to the ground = 14.5 m/s
Final momentum of the system = (70.5 kg * 2.05 m/s) + (0.470 kg * 14.5 m/s)
Final momentum of the system = 144.025 kg·m/s + 6.815 kg·m/s
Final momentum of the system = 150.84 kg·m/s
The final momentum of the system is equal to the total momentum of the player and the ball after the ball is thrown.
Now, let's find the player's speed afterward.
Final momentum of the system = Mass of the system * Final velocity of the system
Given:
Mass of the system = mass of the football player + mass of the thrown ball = 70.5 kg + 0.470 kg = 70.97 kg (approx)
From the equation:
Final momentum of the system = (70.97 kg) * Final velocity of the system
Therefore,
(70.97 kg) * Final velocity of the system = 150.84 kg·m/s
Now, let's solve for the final velocity of the system (player's speed afterward):
Final velocity of the system = 150.84 kg·m/s / 70.97 kg
Final velocity of the system ≈ 2.127 m/s
So, the player's speed afterward, in the first scenario where the ball is thrown at 14.5 m/s relative to the ground, is approximately 2.127 m/s.
Now, let's consider the second scenario where the ball is thrown at 14.5 m/s relative to the player.
The relative velocity of the ball with respect to the player will be the difference between the velocity of the ball relative to the ground and the velocity of the player.
Relative velocity of the ball with respect to the player = Velocity of the thrown ball relative to the ground - Velocity of the player
Given:
Velocity of the thrown ball relative to the ground = 14.5 m/s
Velocity of the player = 2.05 m/s
Relative velocity of the ball with respect to the player = 14.5 m/s - 2.05 m/s
Relative velocity of the ball with respect to the player ≈ 12.45 m/s
So, in the second scenario where the ball is thrown at 14.5 m/s relative to the player, the player's speed afterward will be approximately 12.45 m/s.
To solve these problems, we can use the principle of conservation of momentum. According to this principle, the total momentum before the event (throwing the football) is equal to the total momentum after the event.
Let's start with the first question:
1. What is the player's speed afterward if the ball is thrown at 14.5 m/s relative to the ground?
To solve this, we need to calculate the initial momentum of the system (player + ball) and the final momentum of the system.
Initial momentum (before the throw) = (mass of the player × velocity of the player) + (mass of the ball × velocity of the ball)
Final momentum (after the throw) = (mass of the player × velocity of the player) + (mass of the ball × (velocity of the player + velocity of the ball))
Since the ice is very smooth and there is no external force acting on the system except for the throw, there is no change in momentum of the player before and after the throw. Therefore, the initial momentum is equal to the final momentum.
Initial momentum = Final momentum
(mass of the player × velocity of the player) + (mass of the ball × velocity of the ball) = (mass of the player × final velocity of the system) + (mass of the ball × final velocity of the system)
Now, let's substitute the known values into the equation:
(70.5 kg × 2.05 m/s) + (0.470 kg × 0 m/s) = (70.5 kg + 0.470 kg) × final velocity of the system
After simplifying the equation, we find:
final velocity of the system = (70.5 kg × 2.05 m/s) / (70.5 kg + 0.470 kg)
Calculate the final velocity by dividing the product of the player's mass and velocity by the total mass of the player and the ball:
final velocity of the system = 2.018 m/s
Therefore, the player's speed afterward, when the ball is thrown at 14.5 m/s relative to the ground, is approximately 2.018 m/s.
Now, let's move on to the second question:
2. What is the player's speed afterward if the ball is thrown at 14.5 m/s relative to the player?
In this scenario, the initial momentum and the final momentum equations remain the same as in the previous question. However, this time the velocity of the ball is relative to the player, not the ground.
Substituting the known values into the equation:
(70.5 kg × 2.05 m/s) + (0.470 kg × (-14.5 m/s)) = (70.5 kg + 0.470 kg) × final velocity
After simplifying the equation, we find:
final velocity of the system = (70.5 kg × 2.05 m/s) - (0.470 kg × 14.5 m/s) / (70.5 kg + 0.470 kg)
Calculate the final velocity by subtracting the product of the ball's mass and velocity from the product of the player's mass and velocity, and then dividing by the total mass of the system:
final velocity of the system = 1.924 m/s
Therefore, the player's speed afterward, when the ball is thrown at 14.5 m/s relative to the player, is approximately 1.924 m/s.