Newton's 3 Law of Motion

  • #1
airelemental135
13
0
8. What average force is needed to accelerate a 7.00-gram pellet from rest to 175 m/s over a distance of 0.700 m along the barrel of a rifle?

9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s^2 using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?

10. A 0.140-kg baseball traveling 45.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?3

I tried doing number 8 on my own.

vfinal^2=vinitial^2 + 2ad
175m/s=0^2 + 2a(0.700m)
a= 125m/s
F=ma
F=(7.00g)(125 m/s)
F=875N

Is this correct>? How do you do the others?
 

Answers and Replies

  • #2
stunner5000pt
1,455
2
for number 8 - CAREFUL about the units. Mass is ALWAYS (make it a habit to convert it to) Kilograms.

for number 9 - Well if you are lifting a weight upward by a string, what are the forces that the weight would feel? ANd since this is upward motion you need only to care about those poiting upward and downward. Create a reference system in which upward is positive and downward is negative (the opposite is fine as well, as long as you are consistent with this reference system). Try to form something like this:
Sum of all the forces = Net Force
ANd test means tension here.

For number 10 - What is hte initial velocity, final velocity, distance it traveled before it came to a stop, thus can you find the acceleration? And find the force. Since it is a vector specify the direction, was the force in the original direction that the ball was going or the opposite?
 
  • #3
airelemental135
13
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It says the baseball recoils backward 11.0 cm. Is that the distance?
 
  • #4
stunner5000pt
1,455
2
airelemental135 said:
It says the baseball recoils backward 11.0 cm. Is that the distance?

yes.. and again careful with the units
 
  • #5
airelemental135
13
0
i used the formula vfinal^2=vinitial^2 + 2ad to calculate the velocity, and got 0. Is this correct? if so, the force would end up being 0?
 
  • #6
airelemental135
13
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i meant to calculate the acceleration
 
  • #7
stunner5000pt
1,455
2
hgow did u get zero? What is the value of v1? and the value of distance?
 
  • #8
airelemental135
13
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v initial i put is 45 m/s, v final is 45 m/s distance is .11m
 
  • #9
airelemental135
13
0
what should it be?
 
  • #10
stunner5000pt
1,455
2
airelemental135 said:
v initial i put is 45 m/s, v final is 45 m/s distance is .11m
Read the question. What happens to the ball when it is caught in the met?
 
  • #11
airelemental135
13
0
it bounces back .11 meters
 
  • #12
airelemental135
13
0
oh so vfinal should be 0? because it says its at rest when it hits the mitt
 
  • #13
airelemental135
13
0
Is this right now?

Vfinal ^2 = vinitial ^2 + 2ad
(0)^2=(45.0m/s)62 + 2a(.11m)
0m/s=2025m/s + .22a
a=-9205m/s^2
F=ma
F=(0.140kg)(-9205m/s^2)
F=-1289N

right?
 
  • #14
airelemental135
13
0
are you still there stunner5000pt ?
 

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