Momentum of 1g and 4g Masses: 4:1, root2:1, 1:2, 1:16

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In summary, the homework statement states that two masses of 1g and 4g are moving with equal kinetic energies. The ratio of the masses linear momentum is 1:2.
  • #1
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Homework Statement


two masses of 1g and 4g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is :
a: 4:1
b root2: 1
c; 1:2
d; 1:16

Homework Equations


KE=1/2MV^2
work=FD

The Attempt at a Solution


no idea what to do.. give me a hint please
 
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  • #2
There is yet another formula that you have to use along with the one you've posted. Which do you think it is?
 
  • #3
ProfuselyQuarky said:
There is yet another formula that you have to use along with the one you've posted. Which do you think it is?
pe=ke?
pe=mgh?
 
  • #5
alijan kk said:
pe=ke?
pe=mgh?
None of those, I'm afraid.
alijan kk said:
w= fd?
The problem has nothing to do with work.
 
  • #6
You're looking for the ratio of the magnitudes of the masses linear momentum.
 
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  • #7
ProfuselyQuarky said:
You're looking for the ratio of the magnitudes of the masses linear momentum.
yes ... but what equation or formula i should use...?
 
  • #8
What is the typical equation for momentum?
 
  • #9
ProfuselyQuarky said:
What is the typical equation for momentum?
momentum=mass*velocity
 
  • #10
Yep! That’s the one. Now, we are already given the mass of both objects, so we have to “equal” the energies of them, using the equation in you OP. How do you think you can do that?
 
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  • #11
ProfuselyQuarky said:
Yep! That’s the one. Now, we are already given the mass of both objects, so we have to “equal” the energies of them, using the equation in you OP. How do you think you can do that?
but we don't know velocity?

p=mv
p=1*v

what now? :--
 
  • #12
We've got:

##E_1=E_2##
##\frac {1}{2}(1)v^2=\frac {1}{2}(4)v^2##

What does canceling the 1/2 give you?
 
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  • #13
ProfuselyQuarky said:
We've got:

##E_1=E_2##
##\frac {1}{2}(1)v^2=\frac {1}{2}(4)v^2##

What does canceling the 1/2 give you?
for E1 it is 1/2 or 0.5
and for E2 it is 2 ?

your way of make me understand is awsome!
 
  • #14
Well, we’re not dividing both sides by two, or else, we’d have to divide ##v^{2}_{1}## and ##v^{2}_{2}## by two, as well. We don't want to do that. We’re just canceling out the ½ because it’s on both sides.

That said, what are we left with?
 
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  • #15
1/2=2

that means it is close to option c 1:2 ? and it is the right answer in the book.
thank you..
ProfuselyQuarky said:
Well, we’re not dividing both sides by two, or else, we’d have to divide ##v^{2}_{1}## and ##v^{2}_{2}## by two, as well. We don't want to do that. We’re just canceling out the ½ because it’s on both sides.

That said, what are we left with?
we are left with 1 and 2 ! great am i right because 1/2 is canceled
 
  • #16
Indeed, the answer is 1:2

Without ½ we get that ##(1)v^{2}_{1}=4v^{2}_{2}##. Square both sides and we derive ##v_1=2v_2##. Remember how ##p=mv## give us the momentum? We can now plug in the values.

##p_1=(1)(v_1)##
##p_2=(2)(v_1)##

We’re looking for ratios, thus:

##\frac{p_1}{p_2}=\frac{(1)(v_1)}{(2)(v_1)}##

Canceling out the ##v_1## gives us 1:2
 
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  • #17
ProfuselyQuarky said:
Indeed, the answer is 1:2

Without ½ we get that ##(1)v^{2}_{1}=4v^{2}_{2}##. Square both sides and we derive ##v_1=2v_2##. Remember how ##p=mv## give us the momentum? We can now plug in the values.

##p_1=(1)(v_1)##
##p_2=(2)(v_1)##

We’re looking for ratios, thus:

##\frac{p_1}{p_2}=\frac{(1)(v_1)}{(2)(v_1)}##

Canceling out the ##v_1## gives us 1:2

ohh! great job... thanks
 

1. What is momentum?

Momentum is a measure of an object's mass and velocity. It is calculated by multiplying an object's mass by its velocity.

2. How does the momentum change when the mass changes?

According to the equation p = mv, if the mass of an object increases, its momentum will also increase proportionally. This means that a 4g mass will have four times the momentum of a 1g mass.

3. What is the relationship between the momentum of 1g and 4g masses with a 4:1 ratio?

The momentum of a 4g mass is four times greater than the momentum of a 1g mass in a 4:1 ratio. This is because the ratio of their masses is the same as the ratio of their momentums, since momentum is directly proportional to mass.

4. How does the momentum change when the ratio between masses is 1:2?

If the ratio between masses is 1:2, the momentum of the larger mass will be twice that of the smaller mass. This is because the larger mass has twice the mass of the smaller mass, resulting in double the momentum.

5. What is the significance of the root2:1 ratio in terms of momentum?

The root2:1 ratio means that the larger mass has a momentum that is 1.41 times greater than the smaller mass. This is because the square root of 2 is approximately 1.41. This ratio can be seen in collisions where the masses have different velocities but the same momentum, as the larger mass will have a lower velocity and the smaller mass will have a higher velocity.

Suggested for: Momentum of 1g and 4g Masses: 4:1, root2:1, 1:2, 1:16

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