Work and energy

  • Thread starter alijan kk
  • Start date
  • #1
130
5

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
 
Last edited:

Answers and Replies

  • #2
ProfuselyQuarky
Gold Member
817
527
There is yet another formula that you have to use along with the one you've posted. Which do you think it is?
 
  • #3
130
5
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?
 
  • #6
ProfuselyQuarky
Gold Member
817
527
You're looking for the ratio of the magnitudes of the masses linear momentum.
 
  • Like
Likes alijan kk
  • #7
130
5
You're looking for the ratio of the magnitudes of the masses linear momentum.
yes .... but what equation or formula i should use....?
 
  • #8
ProfuselyQuarky
Gold Member
817
527
What is the typical equation for momentum?
 
  • #10
ProfuselyQuarky
Gold Member
817
527
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?
 
  • Like
Likes alijan kk
  • #11
130
5
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? :--
 
  • #12
ProfuselyQuarky
Gold Member
817
527
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?
 
  • Like
Likes alijan kk
  • #13
130
5
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
ProfuselyQuarky
Gold Member
817
527
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?
 
  • Like
Likes alijan kk
  • #15
130
5
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
 
  • #16
ProfuselyQuarky
Gold Member
817
527
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
 
  • Like
Likes alijan kk
  • #17
130
5
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
 

Related Threads on Work and energy

  • Last Post
Replies
1
Views
557
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
567
  • Last Post
Replies
1
Views
966
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
6
Views
411
  • Last Post
Replies
6
Views
962
  • Last Post
Replies
4
Views
1K
Top