A 0.20 kg baseball moving at +25.20 m/s is slowed to a stop by a catcher who exerts a constant force of -387 N.
A)How long does it take this force to stop the ball?
B)How far does the ball travel before stopping?
F=Δp/Δt =m(v₂-v₁)/Δt
v₂ =0
Δt= m v₂/F=(-0.2•25.2)/(- 387)= 0.013 s
a= - v₁/Δt=-25.2/0.013 = - 1938.5 m/s²
s=- v₁²/2a=-25.2²/2•(-1938.5)=0.16 m
Step 1: Calculate the acceleration of the baseball using Newton's second law, F = ma.
Given:
Mass of the baseball (m): 0.20 kg
Force exerted by the catcher (F): -387 N (negative sign indicates opposite direction)
Acceleration (a) = F / m
a = -387 N / 0.20 kg = -1935 m/s^2
Step 2: Calculate the time taken to stop the ball using the equation of motion, v = u + at, where u is the initial velocity, v is the final velocity, t is the time taken, and a is the acceleration.
Given:
Initial velocity of the baseball (u): 25.20 m/s
Final velocity (v) = 0 m/s
Rearranging the equation, t = (v - u) / a
t = (0 m/s - 25.20 m/s) / -1935 m/s^2
t ≈ 0.013 s
Therefore, it takes approximately 0.013 seconds for the force to stop the ball.
Step 3: Calculate the distance traveled by the ball before stopping using the equation of motion, s = ut + 1/2 at^2.
Given:
Acceleration (a) = -1935 m/s^2
Initial velocity (u) = 25.20 m/s
Time (t) ≈ 0.013 s
Rearranging the equation, s = ut + 1/2 at^2
s = (25.20 m/s)(0.013 s) + 1/2 (-1935 m/s^2)(0.013 s)^2
s ≈ 0.163 m
Therefore, the ball travels approximately 0.163 meters before stopping.
To find the answers to these questions, we need to use the equations of motion. One crucial equation we can use is Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).
A) How long does it take this force to stop the ball?
We know the mass of the baseball (m = 0.20 kg) and the force exerted by the catcher (F = -387 N). We need to find the acceleration of the baseball, which is not given directly. However, we can use Newton's second law to find it.
F = ma
By rearranging the equation, we can solve for acceleration:
a = F / m
Plugging in the given values, we get:
a = -387 N / 0.20 kg
a = -1935 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity of the baseball.
Now, to find the time it takes to stop the ball, we can use another equation of motion:
v = u + at
Where:
v is the final velocity (0 m/s, as the ball stops),
u is the initial velocity (+25.20 m/s),
a is the acceleration we just calculated (-1935 m/s^2), and
t is the time we need to find.
Now, rearranging the equation to solve for time:
t = (v - u) / a
Plugging in the given values:
t = (0 m/s - (+25.20 m/s)) / (-1935 m/s^2)
t = (-25.20 m/s) / (-1935 m/s^2)
t ≈ 0.013 seconds
Therefore, it takes approximately 0.013 seconds for the force to stop the ball.
B) How far does the ball travel before stopping?
To find the distance the ball travels before stopping, we can use the equation:
s = ut + 0.5at^2
Where:
s is the distance traveled (which is what we need to find),
u is the initial velocity (+25.20 m/s),
t is the time we just calculated (0.013 seconds), and
a is the acceleration we calculated earlier (-1935 m/s^2).
Plugging in the values:
s = (+25.20 m/s) * (0.013 s) + 0.5 * (-1935 m/s^2) * (0.013 s)^2
s ≈ 0.163 meters
Therefore, the ball travels approximately 0.163 meters before stopping.