Question

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which the ball is moving back towards the pitcher at 110 mph. The bat was in contact with the ball for 0.0007 sec. The ball weighs 145 g. (There are 1600 meters in a mile.) a. Convert all measurements into MKS units. b. What is the force on the baseball when it is hit by the bat?

Answer 1

Answer:

$F=-18412.9N$, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$, so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$