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

• alijan kk
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.

## 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

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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There is yet another formula that you have to use along with the one you've posted. Which do you think it is?

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?

w= fd?

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.

You're looking for the ratio of the magnitudes of the masses linear momentum.

alijan kk
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...?

What is the typical equation for momentum?

ProfuselyQuarky said:
What is the typical equation for momentum?
momentum=mass*velocity

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?

alijan kk
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? :--

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?

alijan kk
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!

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?

alijan kk
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

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

alijan kk
ProfuselyQuarky said:

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.