A 0.145 kg baseball is moving at a speed of 15.0 meter/sec east toward the bat. The bat has 3.00x10^-3 seconds contact with the ball. As a result the ball flies back at 25.0 meter/sec west. What was the force (vector!) the bat exerted on the ball?
To find the force exerted by the bat on the ball, we can use Newton's Second Law of Motion, which states that force is equal to the change in momentum. The formula for momentum is given by:
Momentum (p) = mass (m) × velocity (v)
In this case, we need to find the change in momentum, which can be calculated as:
Change in momentum = Final momentum - Initial momentum
The initial momentum of the ball can be calculated by multiplying its mass by its initial velocity:
Initial momentum = mass × initial velocity
Final momentum can be calculated by multiplying the mass by the final velocity:
Final momentum = mass × final velocity
Now let's calculate the initial momentum of the ball:
Initial momentum = 0.145 kg × 15.0 m/s (east)
Initial momentum = 2.175 kg·m/s (east)
We also need to calculate the final momentum of the ball:
Final momentum = 0.145 kg × 25.0 m/s (west)
Final momentum = -3.625 kg·m/s (west) [Note: Since the ball is moving in the opposite direction (west), the velocity is considered negative.]
Now, we can calculate the change in momentum:
Change in momentum = Final momentum - Initial momentum
Change in momentum = (-3.625 kg·m/s) - (2.175 kg·m/s)
Change in momentum = -5.800 kg·m/s (west) [Note: Again, since the ball is moving in the opposite direction (west), the value is negative.]
Since force is equal to the change in momentum, the force exerted by the bat on the ball is:
Force = Change in momentum / Time
Force = (-5.800 kg·m/s) / (3.00x10^-3 s)
Force = -1.93333... × 10^6 N (west)
Therefore, the force exerted by the bat on the ball is approximately -1.93 × 10^6 N in the west direction.
To determine the force exerted by the bat on the ball, you can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given:
Mass of the baseball (m) = 0.145 kg
Initial velocity of the baseball (u) = 15.0 m/s (east)
Final velocity of the baseball (v) = -25.0 m/s (west)
Contact time (Δt) = 3.00x10^-3 seconds
First, calculate the change in velocity (Δv) of the baseball:
Δv = v - u
Δv = (-25.0) - (15.0)
Δv = -40.0 m/s
Next, calculate the acceleration (a) of the baseball using the formula:
a = Δv / Δt
a = (-40.0) / (3.00x10^-3)
a = -1.33 x 10^4 m/s^2 (west)
Now, substitute the mass (m) and acceleration (a) into Newton's second law to find the force (F):
F = m * a
F = 0.145 kg * (-1.33 x 10^4 m/s^2)
F ≈ -1.93 x 10^3 N (west)
Therefore, the force exerted by the bat on the ball is approximately -1.93 x 10^3 N (west).