A catcher stops a 200km/h pitch in his glove, bringing it to rest in 0.15meters. If the force exerted by the catcher is 803N, what is the mass of the ball?
v0=200 km/h = 200/3.6 m/s = 55.6 m/s
v1=0
v0^2-v1^2=2aS
S=0.15 m
a=(v0^2-0)/(2S)
=10288 m/s²
F=ma
m=a/F
=10288/803 kg
=0.078 kg.
Unfortunately, baseballs usually weigh around 140 g., so I don't know if something is wrong somewhere. Check my thinking.
Pretty silly question. The official mass of a major league baseball is 0.145 kg. No pitcher has ever thrown a baseball 200 km/h. Not even Sandy Koufax.
You can derive a baseballmass from the numbers you were given by equating the kinetic energy of the baseball to the work done against the glove, and solving for m. What you derive may or may not agree with the actual mass of a baseball.
MathMate had everything right up to a/F. It should be F/a=m
To find the mass of the ball, we can use Newton's Second Law of Motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
In this case, we know the force exerted by the catcher, which is 803N. We also know that the pitch went from a speed of 200km/h to come to rest in the catcher's glove, which means it experienced a deceleration. We need to convert the speed to meters per second (m/s) and calculate the deceleration (a).
1. Convert the speed from km/h to m/s:
- 200 km/h = 200,000 m/60 mins (since there are 60 minutes in an hour) = 3,333.33 m/min
- 3,333.33 m/min ÷ 60s (since there are 60 seconds in a minute) = 55.56 m/s
2. Calculate the deceleration (a):
- The pitch comes to rest, so the final velocity (vf) is 0 m/s, and the initial velocity (vi) is 55.56 m/s.
- Using the formula for acceleration (a = (vf - vi) / t) and assuming the time it took to stop is 1 second (to simplify the calculations):
a = (0 m/s - 55.56 m/s) / 1 s = -55.56 m/s^2
3. Substitute the values into Newton's Second Law of Motion:
F = m * a
803N = m * (-55.56 m/s^2)
4. Solve for mass (m):
m = 803N / (-55.56 m/s^2)
m ≈ - 14.44 kg
Since mass cannot be negative, we conclude that there has been an error in the calculations. Please double-check the given values and calculations to find the correct mass of the ball.