Work and energy

1. May 29, 2016

alijan kk

1. The problem statement, all variables and given/known data
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

2. Relevant equations
KE=1/2MV^2
work=FD

3. The attempt at a solution
no idea what to do.. give me a hint please

Last edited: May 29, 2016
2. May 29, 2016

ProfuselyQuarky

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

3. May 29, 2016

alijan kk

pe=ke?
pe=mgh?

4. May 29, 2016

alijan kk

w= fd?

5. May 29, 2016

ProfuselyQuarky

None of those, I'm afraid.
The problem has nothing to do with work.

6. May 29, 2016

ProfuselyQuarky

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

7. May 29, 2016

alijan kk

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

8. May 29, 2016

ProfuselyQuarky

What is the typical equation for momentum?

9. May 29, 2016

alijan kk

momentum=mass*velocity

10. May 29, 2016

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?

11. May 29, 2016

alijan kk

but we dont know velocity?

p=mv
p=1*v

what now? :--

12. May 29, 2016

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?

13. May 29, 2016

alijan kk

for E1 it is 1/2 or 0.5
and for E2 it is 2 ?

your way of make me understand is awsome!!

14. May 29, 2016

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?

15. May 29, 2016

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..
we are left with 1 and 2 !!!!! great am i right because 1/2 is canceled

16. May 29, 2016

ProfuselyQuarky

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

17. May 29, 2016

alijan kk

ohh! great job... thanks