Simple and basic momentum and work done Qs!

  • #1
nonthesecond
22
0
Mr Green, mass 80 kg, stepped off a boat of mass 35 kg at a speed of 3m/s. how fast did the boat move in the opposite direction?
 

Answers and Replies

  • #2
rollcast
409
0
As this sounds like a homework problems I'll just give you a few hints abut the problem.

Conservation of momentum for the total system states that the momentum before and after must be equal as there are no external forces acting upon the system.

So before he steps off as I'm assuming the boat is at rest, the total momentum of the system will be 0.

So after the man steps off the boat his momentum will be equal and opposite to that of the boat as the 2 momentums will cancel out to be zero.
 
  • #3
nonthesecond
22
0
As this sounds like a homework problems I'll just give you a few hints abut the problem.

Conservation of momentum for the total system states that the momentum before and after must be equal as there are no external forces acting upon the system.

So before he steps off as I'm assuming the boat is at rest, the total momentum of the system will be 0.

So after the man steps off the boat his momentum will be equal and opposite to that of the boat as the 2 momentums will cancel out to be zero.

thanks which formulae do i use?

m = mass x v

or

mom before = mom after

?
 
  • #4
rollcast
409
0
You can use either but probably working out the momentum of the man using p = m * v, then use that value of p to work out the momentum of the boat, v = m / p. If you were working out the velocity of the boat afterwards you would need to make momentum negative but as it only asks for the speed you can keep momentum positive.
 
  • #5
nonthesecond
22
0
You can use either but probably working out the momentum of the man using p = m * v, then use that value of p to work out the momentum of the boat, v = m / p. If you were working out the velocity of the boat afterwards you would need to make momentum negative but as it only asks for the speed you can keep momentum positive.

is this correct?

80 x 3 = 240

80/240 = 0.3 (which is the velocity) so it's 0.3m/s^2
 
  • #6
nonthesecond
22
0
You can use either but probably working out the momentum of the man using p = m * v, then use that value of p to work out the momentum of the boat, v = m / p. If you were working out the velocity of the boat afterwards you would need to make momentum negative but as it only asks for the speed you can keep momentum positive.

please help
 
  • #7
rollcast
409
0
is this correct?

80 x 3 = 240

80/240 = 0.3 (which is the velocity) so it's 0.3m/s^2

Sorry I made a mistake rearranging the formula for velocity.

It should be v=p/m

Your calculation for the momentum is correct and you should get the right answer if you use the right formula for the velocity.
 
  • #8
nonthesecond
22
0
Sorry I made a mistake rearranging the formula for velocity.

It should be v=p/m

Your calculation for the momentum is correct and you should get the right answer if you use the right formula for the velocity.

how can it be 3m/s^2 because that ids already given in the question.

i think i remeber my teacher saying that it's: 80 + 35= 115
115 x 3 = 375
375/35 = 1.3...m/s

is that correct?
 
  • #9
rollcast
409
0
how can it be 3m/s^2 because that ids already given in the question.

i think i remeber my teacher saying that it's: 80 + 35= 115
115 x 3 = 375
375/35 = 1.3...m/s

is that correct?

Thats how you would calculate the initial momentum for the system as the 2 masses are together and have the same velocity.

Initial momentum for the system = m * v = (80 + 35) * 0


Then you need to treat the 2 masses separate for the after stage as the 2 masses are not together anymore and will be moving in opposite directions (Newtons 3rd Law)
 
  • #10
nonthesecond
22
0
Thats how you would calculate the initial momentum for the system as the 2 masses are together and have the same velocity.

Initial momentum for the system = m * v = (80 + 35) * 0


Then you need to treat the 2 masses separate for the after stage as the 2 masses are not together anymore and will be moving in opposite directions (Newtons 3rd Law)


ok thanks so what would i have to do to find how fast the boat moved in the opposite direction?
 
  • #11
rollcast
409
0
ok thanks so what would i have to do to find how fast the boat moved in the opposite direction?

Work out the momentum for Mr Green.

p = m * v

Then use this value as the momentum for the boat (You can ignore the fact it should be opposite as we are only working out speed not velocity) Rearrange the momentum formula to work out velocity (or speed in this case).

p = m * v
v = p / m
 
  • #12
nonthesecond
22
0
Work out the momentum for Mr Green.

p = m * v

Then use this value as the momentum for the boat (You can ignore the fact it should be opposite as we are only working out speed not velocity) Rearrange the momentum formula to work out velocity (or speed in this case).

p = m * v
v = p / m

3 m/s is it squared but how can that be the answer when it's already written in the question?
 
  • #13
rollcast
409
0
3 m/s is it squared but how can that be the answer when it's already written in the question?

For the second step you need to use the mass of the boat.
 
  • #14
nonthesecond
22
0
For the second step you need to use the mass of the boat.

so it's 240/35 ? = 6.9m/s is that the final answer and it's not squared?
 
  • #15
nonthesecond
22
0
For the second step you need to use the mass of the boat.

did i get the question right?
 

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