A 0.18 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of −360 N.
a) How long does it take this force to stop the ball?
b) How far does the ball travel before stop- ping?
force * time = change in momentum
360 * t = .18*26
solve for t
a = change in velocity/time
= -26/t
Vi = 26
d = Vi t + (1/2)a t^2
d = 26 t - (1/2)(26)t
d = 13 t
To solve this problem, we can use the equations of motion. The equation that relates force, mass, acceleration, and time is:
F = ma
Where:
F is the force applied,
m is the mass of the object,
a is the acceleration of the object.
In this case, the force applied by the catcher is constant and is equal to -360 N. The mass of the baseball is 0.18 kg.
a) How long does it take this force to stop the ball?
To find the time taken to stop the ball, we need to calculate the acceleration first.
From the equation F = ma, we can rearrange it to solve for acceleration:
a = F/m
a = (-360 N) / (0.18 kg)
a = -2000 m/s^2
Since the initial velocity is +26 m/s and the final velocity is 0 m/s when the ball stops, we can use the following equation to find the time taken:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration of the object,
t is the time taken.
From the equation, we can solve for t:
0 = 26 m/s + (-2000 m/s^2) * t
Rearranging the equation, we have:
26 m/s = (-2000 m/s^2) * t
t = 26 m/s / 2000 m/s^2
t = 0.013 s
Therefore, it takes about 0.013 seconds for the catcher to stop the ball.
b) How far does the ball travel before stopping?
To find the distance traveled, we can use the following equation:
s = ut + 0.5at^2
Where:
s is the distance traveled,
u is the initial velocity,
t is the time taken,
a is the acceleration of the object.
Substituting the given values:
s = (26 m/s) * (0.013 s) + 0.5 * (-2000 m/s^2) * (0.013 s)^2
Simplifying the equation:
s = 0.338 m
Therefore, the ball travels about 0.338 meters before stopping.
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. We can use this equation to find the acceleration of the baseball:
F = m * a
Where:
F is the force applied by the catcher (-360 N)
m is the mass of the baseball (0.18 kg)
a is the acceleration of the baseball
a) To find the acceleration:
Rearrange the equation to solve for acceleration:
a = F / m
Substitute the given values:
a = -360 N / 0.18 kg
Calculate the acceleration:
a = -2000 m/s² (since force and acceleration have opposite signs)
Next, we can use the kinematic equation to find the time it takes for the baseball to stop:
v = u + at
Where:
v is the final velocity of the baseball (0 m/s, since it comes to a stop)
u is the initial velocity of the baseball (+26 m/s)
a is the acceleration of the baseball (-2000 m/s²)
t is the time taken to stop
b) To find the time taken to stop:
Substitute the given values:
0 = 26 m/s + (-2000 m/s²) * t
Rearrange the equation and solve for time:
t = (0 - 26 m/s) / -2000 m/s²
Calculate the time taken to stop:
t = 0.013 seconds (rounded to three decimal places)
Now, to find the distance traveled before stopping, we can use another kinematic equation:
s = ut + (1/2)at²
Where:
s is the distance traveled before stopping
u is the initial velocity of the baseball (+26 m/s)
t is the time taken to stop (0.013 seconds)
a is the acceleration of the baseball (-2000 m/s²)
Substitute the given values:
s = 26 m/s * 0.013 s + (1/2) * (-2000 m/s²) * (0.013 s)²
Calculate the distance traveled before stopping:
s = 0.170 meters (rounded to three decimal places)
Therefore, the answers to the questions are:
a) It takes 0.013 seconds for the catcher to stop the ball.
b) The ball travels approximately 0.170 meters before stopping.