A 0.17 kg baseball moving at +20.80 m/s is slowed to a stop by a catcher who exerts a constant force of -376 N.
How long does it take this force to stop the ball?
How far does the ball travel before stopping?
Well, stopping a baseball is no easy task, especially when it's moving at such a speed. But don't worry, I'm here to calculate it for you!
To find out how long it takes for the ball to stop, we can use Newton's second law, which says that force equals mass times acceleration (F = ma). In this case, the force is -376 N and the mass is 0.17 kg. So we can rearrange the equation to solve for acceleration: a = F/m.
a = -376 N / 0.17 kg
a ≈ -2211.76 m/s²
Since the force is constant and in the opposite direction to the motion of the ball, the acceleration is negative. Now, to find the time it takes for the ball to stop, we use the formula for motion with constant acceleration: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
0 = 20.80 m/s + (-2211.76 m/s²)t
t ≈ 20.80 m/s ÷ 2211.76 m/s²
t ≈ 0.0094 s
So it takes approximately 0.0094 seconds for the ball to come to a stop.
Now, let's figure out how far the ball travels before stopping. We can use another equation of motion: s = ut + 0.5at², where s is the distance traveled. Since the final velocity is 0 m/s, we can rearrange the formula to solve for distance:
s = (20.80 m/s)(0.0094 s) + 0.5(-2211.76 m/s²)(0.0094 s)²
s ≈ 0.1845 m
Therefore, the ball travels approximately 0.1845 meters before coming to a complete stop. That's quite a distance for a catcher to catch!
To find the time it takes for the force to stop the ball, we can use the equation:
force = mass * acceleration
The acceleration can be calculated using Newton's second law:
acceleration = force / mass
Given that the mass of the baseball is 0.17 kg and the force exerted by the catcher is -376 N (assuming "positive" direction is opposite to the direction of motion), we can substitute these values into the equation:
acceleration = -376 N / 0.17 kg
Using this equation, we can calculate the acceleration:
acceleration = -2211.76 m/s^2
Next, we can use the kinematic equation to find the time it takes for the ball to stop, assuming it starts from a speed of +20.80 m/s:
final velocity = initial velocity + acceleration * time
We know that the final velocity is 0 m/s (since the ball stops), the initial velocity is +20.80 m/s, and the acceleration is -2211.76 m/s^2. Substituting these values into the equation, we can solve for time:
0 = 20.80 m/s - 2211.76 m/s^2 * time
Simplifying the equation, we have:
20.80 m/s = 2211.76 m/s^2 * time
Dividing both sides by 2211.76 m/s^2, we get:
time = 20.80 m/s / 2211.76 m/s^2
Calculating this, we find:
time ≈ 0.0094 s
Therefore, it takes approximately 0.0094 seconds for the force to stop the ball.
To find the distance the ball travels before stopping, we can use the formula for distance:
distance = initial velocity * time + 0.5 * acceleration * time^2
Since the final velocity is 0 m/s (the ball stops), we can simplify the equation to:
distance = initial velocity * time
Substituting in the values, we get:
distance = 20.80 m/s * 0.0094 s
Calculating this, we find:
distance ≈ 0.1952 m
Therefore, the ball travels approximately 0.1952 meters before stopping.
To find the time it takes for the force to stop the ball, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. The formula can be rearranged to solve for time (t):
a = F/m
Where:
a = acceleration of the baseball (which in this case is the negative value of deceleration, as the ball is slowing down)
F = force exerted by the catcher (-376 N)
m = mass of the baseball (0.17 kg)
t = time taken to stop the ball
First, let's calculate the acceleration of the baseball:
a = F/m
a = (-376 N) / (0.17 kg)
a ≈ -2211.76 m/s^2
Now, we have the acceleration of the baseball. Since the initial velocity of the baseball is +20.8 m/s and it comes to a stop (final velocity is 0 m/s), we can use the formula of motion to find the time taken to stop:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (20.8 m/s)
a = acceleration (-2211.76 m/s^2)
t = time
Rearranging the formula:
t = (v - u) / a
t = (0 m/s - 20.8 m/s) / (-2211.76 m/s^2)
t ≈ 0.0094 s (rounded to four decimal places)
Therefore, the force takes approximately 0.0094 seconds to stop the ball.
To find the distance the ball travels before stopping, we can use the formula for displacement:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (20.8 m/s)
t = time (0.0094 s)
a = acceleration (-2211.76 m/s^2)
Plugging in the values:
s = (20.8 m/s) * (0.0094 s) + (1/2) * (-2211.76 m/s^2) * (0.0094 s)^2
s ≈ 0.0978 m (rounded to four decimal places)
Therefore, the ball travels approximately 0.0978 meters before stopping.
Force*Time = Initial Momentum = M*V
Time = (0.17)*(20.8)/376 = 9.4*10^-3 s
(Initial velocity)/2 * Time
= Stopping Distance
= 10.40*9.4*10^-3 = 0.0978 m
= 9.7 cm