The momentum and the kinetic energy

[Asmaa Mohammad]

Homework Statement

Multiple choice question:
The momentum of a body has increased by 25%, then its kenitic energy will roughly increase by...
1. 25% 2. 5% 3. 38% 4. 56% 5. 65%

Homework Equations

Pl = mv (Pl is the momentum, m is mass and v is velocity)
K.E = 0.5 mv^2

The Attempt at a Solution

I don't know how to start, so I will appreciate some hints.

 

[gneill]

Start by completing the template... what are the relevant equations that pertain to the problem?

 

[BvU]

Do you have expressions for momentum and kinetic energy ?
List them under "2. relevant equations" !

 

[Asmaa Mohammad]

Start by completing the template... what are the relevant equations that pertain to the problem?

Do you have expressions for momentum and kinetic energy ?
List them under "2. relevant equations" !

I added the equations, then what?

 

[BvU]

Take a body; let its mass be $m$. If it has a speed $v$, the momentum will be $mv$ and the kinetic energy will be ${1\over 2} mv^2$, right ?
Now increase its momentum to $p'$ with $p'= 1.25\, p$. It's still the same body, so the $m$ remains the same. What changes is $v$ -- let the new value be $v'$. Now work out ${1\over 2} mv'^{\, 2} / {1\over 2} mv^ 2 \$.
That's all.

 

[Asmaa Mohammad]

Take a body; let its mass be $m$. If it has a speed $v$, the momentum will be $mv$ and the kinetic energy will be ${1\over 2} mv^2$, right ?
Now increase its momentum to $p'$ with $p'= 1.25\, p$. It's still the same body, so the $m$ remains the same. What changes is $v$ -- let the new value be $v'$. Now work out ${1\over 2} mv'^{\, 2} / {1\over 2} mv^ 2 \$.
That's all.

OK, I got 0.64. That means the percentage will be 64%, and the correct answer is 65%. Right?

 

[BvU]

No. Show your steps, not just he answer.

And: it would be strange if the kinetic energy went from 100 % to 64 % (how do you get 0.64 ?)

 

[cnh1995]

OK, I got 0.64. That means the percentage will be 64%, and the correct answer is 65%. Right?

No. What is the new velocity if the original velocity was v?

 

[Asmaa Mohammad]

No. What is the new velocity if the original velocity was v?

No. Show your steps, not just he answer.

And: it would be strange if the kinetic energy went from 100 % to 64 % (how do you get 0.64 ?)

Ok, here is my steps, where is the mistake?

 

[cnh1995]

Ok, here is my steps, where is the mistake?
View attachment 205154

Initially, the momentum was p and the velocity was v. Now the momentum is 1.25p, with mass unchanged. So what is the new velocity?

 

[Asmaa Mohammad]

Initially, the momentum was p and the velocity was v. Now the momentum is 1.25p, with mass unchanged. So what is the new velocity?

The new velocity is
v = 1.25p/ m
isn't it?

 

[cnh1995]

The new velocity is
v = 1.25p/ m
isn't it?

What is the new velocity in terms of v?
Or use actual numbers.
Initial momentum=1, m=1, v=1. Now the momentum is 1.25 with m still equal to 1. What is the new velocity now?

 

[Asmaa Mohammad]

What is the new velocity in terms of v?
Or use actual numbers.
Initial momentum=1, m=1, v=1. Now the momentum is 1.25 with m still equal to 1. What is the new velocity now?

the new velocity (v2) is (1.25v1) where v1 is the original velocity, right?
Or if you want it just in numbers it will be 1.25 since m is still equal to 1, right?

 

[BvU]

the new velocity (v2) is (1.25v1) where v1 is the original velocity, right?
Or if you want it just in numbers it will be 1.25 since m still equal to 1, right?

The original question asks for KE₂ / KE₁ x 100 % - 100 % (the increase in KE in percent)

You had the right answer (1.56 * 100 % - 100 % ) but handled it the wrong way around !

Case solved, onto the next exercise !

 

[cnh1995]

the new velocity (v2) is (1.25v1) where v1 is the original velocity, right?
Or if you want it just in numbers it will be 1.25 since m still equal to 1, right?

Yes. Even if m is not 1, still v2=1.25v1.

Now compute kinetic energy E2. You know E1=1.

 

[Asmaa Mohammad]

The original question asks for KE2 / KE1 x 100 % - 100 % (the increase in KE in percent)

I don't understand this line, why did you interpret the question as this?

 

[cnh1995]

Now compute kinetic energy E2. You know E1=1.

 

[Asmaa Mohammad]

@cnh1995
E2 = 0.5 m (1.25v)^2 = 0.5 m (1.5625 v^2) = 1.5625 E1
Right?

 

[cnh1995]

@cnh1995
E2 = 0.5 m (1.25v)^2 = 0.5 m (1.5625 v^2) = 1.5625 E1
Right?

Right.

 

[Asmaa Mohammad]

Right.

Ok, how could I get the percentage and choose the correct answer?

 

[cnh1995]

Ok, how could I get the percentage and choose the correct answer?

What is the 'increase' in energy? What percentage of E1 is that?

 

[Asmaa Mohammad]

Oh, sorry I made a mistake, a very silly one. It will be 0.5625 and when we multiply it to 100 it will be 56.25, so the percentage will be 56% and that's the correct answer. My apologies.

 

[cnh1995]

Oh, sorry I made a mistake, a very silly one. It will be 0.5625 and when we multiply it to 100 it will be 56.25, so the percentage will be 56% and that's the correct answer. My apologies.

I don't think that's a mistake. E2=1.5625*E1 is correct. This is a 56% 'increase'.

Anyways, you can now mark it solved (and I can be off to bed)!

 

[Asmaa Mohammad]

I don't think that's a mistake. E2=1.5625*E1 is correct. This is a 56% 'increase'.

It was a mistake in a previous thread, but it is now deleted.

Anyways, you can now mark it solved (and I can be off to bed)!

Good night !