# Work and energy

## 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

Last edited:

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

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?

ProfuselyQuarky
Gold Member
pe=ke?
pe=mgh?
None of those, I'm afraid.
w= fd?
The problem has nothing to do with work.

ProfuselyQuarky
Gold Member
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
Gold Member
What is the typical equation for momentum?

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

ProfuselyQuarky
Gold Member
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 dont know velocity?

p=mv
p=1*v

what now? :--

ProfuselyQuarky
Gold Member
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
Gold Member
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
Gold Member

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

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