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

[alijan kk]

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

 

[ProfuselyQuarky]

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

 

[alijan kk]

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?

 

[alijan kk]

w= fd?

 

[ProfuselyQuarky]

pe=ke?
pe=mgh?

None of those, I'm afraid.

w= fd?

The problem has nothing to do with work.

 

[ProfuselyQuarky]

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

 

[alijan kk]

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

yes ... but what equation or formula i should use...?

 

[ProfuselyQuarky]

What is the typical equation for momentum?

 

[alijan kk]

What is the typical equation for momentum?

momentum=mass*velocity

 

[ProfuselyQuarky]

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]

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? :--

 

[ProfuselyQuarky]

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]

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!

 

[ProfuselyQuarky]

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..

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

 

[ProfuselyQuarky]

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

 

[alijan kk]

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