# The momentum and the kinetic energy

Tags:
1. Jun 9, 2017

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

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

3. The attempt at a solution
I don't know how to start, so I will appreciate some hints.

Last edited: Jun 9, 2017
2. Jun 9, 2017

### Staff: Mentor

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

3. Jun 9, 2017

### BvU

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

4. Jun 9, 2017

I added the equations, then what?

5. Jun 9, 2017

### 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.

6. Jun 9, 2017

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

7. Jun 9, 2017

### BvU

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

8. Jun 9, 2017

### cnh1995

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

9. Jun 9, 2017

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

10. Jun 9, 2017

### cnh1995

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?

11. Jun 9, 2017

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

12. Jun 9, 2017

### cnh1995

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?

13. Jun 9, 2017

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?

14. Jun 9, 2017

### BvU

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 !

15. Jun 9, 2017

### cnh1995

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

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

16. Jun 9, 2017

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

17. Jun 9, 2017

### cnh1995

18. Jun 9, 2017

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

19. Jun 9, 2017

### cnh1995

Right.

20. Jun 9, 2017