An inexperienced catcher catches a 128 km/h fastball of mass 134 g within 1.01 ms, whereas an experienced catcher slightly retracts his hand during the catch, extending the stopping time to 10.1 ms. What is the average force imparted to the two gloved hands during the catches of the inexperienced catcher?

What is the average force imparted to the two gloved hands during the catches of the experienced catcher?

To find the average force imparted to the two gloved hands during the catches, we can use the equation:

Average force = Change in momentum / Time

First, let's calculate the change in momentum for each catcher.

For the inexperienced catcher:
Initial momentum = mass * initial velocity
Final momentum = mass * final velocity

The change in momentum is given by:
Change in momentum = Final momentum - Initial momentum

Initial momentum = (mass * initial velocity) = (0.134 kg) * (128 km/h) = (0.134 kg) * ((128 km/h) * (1000 m/1 km) * (1 h/3600 s)) = 4.98 kg m/s

Final velocity = 0 m/s (as the ball is caught and brought to a stop)
Final momentum = (mass * final velocity) = (0.134 kg) * (0 m/s) = 0 kg m/s

Change in momentum = Final momentum - Initial momentum = 0 kg m/s - 4.98 kg m/s = -4.98 kg m/s

Next, let's calculate the average force for the inexperienced catcher:
Average force = Change in momentum / Time = (-4.98 kg m/s) / (1.01 ms) = (-4.98 kg m/s) / (0.00101 s) = -4920.79 N (approximately)

Now, let's do the same calculations for the experienced catcher:

Change in momentum = Final momentum - Initial momentum

Initial momentum = (mass * initial velocity) = (0.134 kg) * (128 km/h) = 4.98 kg m/s
Final velocity = 0 m/s (as the ball is caught and brought to a stop)
Final momentum = (mass * final velocity) = (0.134 kg) * (0 m/s) = 0 kg m/s

Change in momentum = Final momentum - Initial momentum
= 0 kg m/s - 4.98 kg m/s
= -4.98 kg m/s

Average force = Change in momentum / Time
= (-4.98 kg m/s) / (10.1 ms)
= (-4.98 kg m/s) / (0.0101 s)
= -492.07 N (approximately)

Therefore, the average force imparted to the two gloved hands during the catches of the inexperienced catcher is approximately -4920.79 N, and the average force imparted to the two gloved hands during the catches of the experienced catcher is approximately -492.07 N.

To find the average force imparted to the two gloved hands during the catches, we can use Newton's second law of motion which states that the force acting on an object is equal to the rate of change of its momentum.

The momentum of an object is given by the product of its mass and velocity. So we need to first calculate the momentum of the fastball.

1. Convert the mass of the fastball from grams to kilograms:
The mass of the fastball is 134 grams.
1 kilogram (kg) = 1000 grams.
Therefore, the mass of the fastball is 134 grams / 1000 grams/kg = 0.134 kg.

2. Convert the velocity of the fastball from km/h to m/s:
The velocity of the fastball is 128 km/h.
1 kilometer (km) = 1000 meters (m).
1 hour (h) = 3600 seconds (s).
Therefore, the velocity of the fastball is 128 km/h * (1000 m/km) / (3600 s/h) = 35.555 m/s (rounded to three decimal places).

Now we can calculate the momentum of the fastball:
Momentum = mass * velocity
Momentum = 0.134 kg * 35.555 m/s = 4.785 kg·m/s (rounded to three decimal places).

To calculate the average force imparted to the gloved hands during the catch, we need to divide the change in momentum by the change in time.

For the inexperienced catcher:
Change in momentum = Final momentum - Initial momentum
= 0 - 4.785 kg·m/s (since the fastball comes to a stop)
= -4.785 kg·m/s

Change in time = 1.01 ms
= 1.01 * 10^-3 s

Average force = Change in momentum / Change in time
= (-4.785 kg·m/s) / (1.01 * 10^-3 s)
= -4735.643 N

Note that the force is negative because it is in the opposite direction of the fastball's velocity.

Therefore, the average force imparted to the gloved hands during the catch of the inexperienced catcher is approximately -4735.643 N.

Now let's calculate the average force for the experienced catcher who retracts his hand and extends the stopping time to 10.1 ms.

For the experienced catcher:
Change in momentum = Final momentum - Initial momentum
= 0 - 4.785 kg·m/s (since the fastball comes to a stop)
= -4.785 kg·m/s

Change in time = 10.1 ms
= 10.1 * 10^-3 s

Average force = Change in momentum / Change in time
= (-4.785 kg·m/s) / (10.1 * 10^-3 s)
= -473.267 N

Therefore, the average force imparted to the gloved hands during the catch of the experienced catcher is approximately -473.267 N.