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klemol [59]
2 years ago
13

A baseball with a mass of 151 g is thrown horizontally with a speed of 39.5 m/s (88 mi/h) at a bat. The ball is in contact with

the bat for 1.10 ms and then travels straight back at a speed of 45.1 m/s (101 mi/h). Determine the average force exerted on the ball by the bat.
Physics
1 answer:
Oduvanchick [21]2 years ago
7 0

Answer:

the average force exerted on the ball by the bat is 11,613.27 N

Explanation:

Given;

mass of the baseball, m = 151 g = 0.151 kg

initial velocity of the baseball, u = 39.5 m/s

final velocity of the baseball, v = 45.1 m/s

time of action, t = 1.10 ms = 1.10 x 10⁻³ s

The average force exerted on the ball by the bat is calculate as;

F = ma = \frac{m(v-u)}{t} \\\\F = \frac{0.151(45.1-(-39.5))}{1.10\times 10^{-3}} \\\\F = \frac{0.151(45.1\ +\ 39.5)}{1.10\times 10^{-3}} \\\\F = 11,613.27 \ N

Therefore, the average force exerted on the ball by the bat is 11,613.27 N

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sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question.  There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems.   That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick.  The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.  

When the question talks about hanging a weight, it tells the <em>centimeter mark on the stick</em> where the weight is tied.  To solve the problem, we have to first figure out <em>how far that is from the pivot</em>, then calculate how far from the pivot to put the weight on the other side, and finally <u><em>what centimeter mark that is</em></u> on the stick.      

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =  

#1).  Only one moment on the left side.  

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm)  =  9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot.  But that's not the answer to the problem.  They want to know what mark on the stick to hang it from.  The pivot is at the 50cm mark.  The 500gm has to hang 9.6 cm to the right of the pivot.  That's the <em>59.6 cm</em> mark on the stick.

= = = = =

#2).  There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight:  120gm, hanging from the 25cm mark.

That's 25cm from the pivot.  Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight:  20gm, hanging from the 10cm mark;

That's 40cm from the pivot.  Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick.  You have to hang the 500gm from 7.6cm to the right of that.  The mark at that spot on the stick is                (50cm + 7.6cm) = <em>57.6 cm </em>.

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