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Force and recoil problem.

  • Thread starter vac
  • Start date
  • #1
vac
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Homework Statement


Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock from rest until he releases it is 1.0 m.

Homework Equations


a. What constant force must Bob exert on the rock to throw it with this speed?
b. If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

The Attempt at a Solution



Known
vi = 0
vf = 30 m/s
delta x = 1 m

[itex]v^2 = vi^2 + 2ax = a = 450 m/s^2[/itex]
[itex]F_{net} = m_{total} * a = (75 kg + 0.500 kg)*450 m/s^2 = 33975 N[/itex]

recoil have no idea.
 

Answers and Replies

  • #2
199
15
[itex]F_{net} = m_{total} * a = (75 kg + 0.500 kg)*450 m/s^2 = 33975 N[/itex]
You'll want to check that again!!!...its not correct. Acceleration is right.

For recoil conserve momentum.
 
  • #3
vac
28
0
You'll want to check that again!!!...its not correct. Acceleration is right.

For recoil conserve momentum.
I know it is not correct, can you please show me the mistake I have done?
 
  • #4
199
15
I know it is not correct, can you please show me the mistake I have done?
you have taken mass, incorrectly. What is being accelerated????
 
  • #5
vac
28
0
For the rock is this 0.500kg(450m/s^2) = 225 N
But the other force I can't get correctly.
 
  • #6
199
15
For the rock is this 0.500kg(450m/s^2) = 225 N
But the other force I can't get correctly.
You don't have to!!!...I hope you have studied momentum conservation. because the force applied is internal to the system (Bob+rock), their momentum is conserved before and after the releasing the rock. I hope this makes it clear!
 
  • #7
vac
28
0
You don't have to!!!
I hate to disappoint you but I actually have to.
The textbook answer is: 2.3x10^2 N = 230 N
I just don't know how to get that.

Now I am guessing that they round it up just to through me in a loop.
 
  • #8
199
15
I hate to disappoint you but I actually have to.
The textbook answer is: 2.3x10^2 N = 230 N
I just don't know how to get that.

Now I am guessing that they round it up just to through me in a loop.
for part (a), you got the right answer....its just rounded off ##225≈230##, because you only have two significant figures.

part (b), you need to calculate velocity not force, velocity can be found out easily be applying conservation of momentum

if its something else, I really don't know what you are talking about!!!:rolleyes:

Edit: I guess you meant to find some other force which was applying 5 Newtons. You'll have learn about this, in almost every other scientific/science calculation there is rounding off. Read about significant figures and scientific notation to learn more.
 
  • #9
vac
28
0
Yes I have, I actually took three semester chemistry and significant figures date back to algebra. How do you apply this conservation of momentum? I know that I studied it before but I just don't remember how?
 
  • #10
199
15
Yes I have, I actually took three semester chemistry and significant figures date back to algebra. How do you apply this conservation of momentum? I know that I studied it before but I just don't remember how?
The formula is,

##m_{1}.v_{1i}+m_{2}.v_{2i}=m_{1}.v_{1f}+m_{2}.v_{2f}##

You just have to fill in the data, since Bob is on frictionless ice, no horizontal external force would act on system (Bob+rock). ask yourself, what was the velocity of rock and Bob before the rock was thrown??
 

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