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Mechanics}Conservation of Momentum

From conservation of momentum, one has 2mv = 3mv' \Rightarrow v'=2/3 v.

The initial kinetic energy is 1/2 m v^2 = mv^2.

The final kinetic energy is 1/2 (3m) v'^2 = 3/2 m (2/3 v)^2.

The ratio of the final to initial kinetic energy is 2/3.

The kinetic energy lost in the collision is 1 minus that ratio. Thus, the answer is 1/3, as in choice (C).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
amber
2014-10-22 15:11:44
Look at the problem thru Center of Mass

Again find the velocity of CM which equals the initial and final velocities of each mass

V = m1v1/ m1+m2 = 2v1/3 = v_final

set the ratio KEf/KEi

0.5 3m 4v1^2/9 / 0.5 2m v1^2 = 2/3

Then 1 - 2/3 to get the energy lost 1/3
Alternate Solution - Unverified
secretempire1
2012-08-28 06:47:39
Formula for kinetic energy K=\frac{1}{2} mv^2

Let m_{1} and v_{1} correspond to the mass and velocity of block 1, m_{2} and v_{2} for block 2, and m_{f} and v_{f} for the resultant combined mass block, where m_{f} = 3m.

Total initial kinetic energy K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2

Since v_{2}=0, K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2

Total final kinetic energy K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2

Performing the same operation for momentum:

m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f}

Solving for v_{f} will yield v_{f} = \frac{2}{3}v_{1}

Plug this into the equation for final kinetic energy, and you get K_{f} = \frac{2}{3}mv_{1}^2

This implies that 1/3 of the energy is lost.
Alternate Solution - Unverified
Comments
Nebula
2015-09-18 20:50:37
Don\'t misread this question and think they are asking for the ratio of final kinetic energy to initial kinetic energy, in which you would get answer (E) 2/3. Easy points lost.NEC
amber
2014-10-22 15:11:44
Look at the problem thru Center of Mass

Again find the velocity of CM which equals the initial and final velocities of each mass

V = m1v1/ m1+m2 = 2v1/3 = v_final

set the ratio KEf/KEi

0.5 3m 4v1^2/9 / 0.5 2m v1^2 = 2/3

Then 1 - 2/3 to get the energy lost 1/3
Alternate Solution - Unverified
Dax Feliz
2012-11-06 22:03:16
Does any one know why we subtract the ratio from 1 to get the fractional loss of energy? I think I missed something here...NEC
CortunaMJM
2012-10-09 02:40:13
how to get the initial kinetic energy if there's no velocity given?
ezfzx
2012-10-13 17:48:25
Since the problem is only looking for a RATIO of energy, the velocity is not needed.
Answered Question!
secretempire1
2012-08-28 06:47:39
Formula for kinetic energy K=\frac{1}{2} mv^2

Let m_{1} and v_{1} correspond to the mass and velocity of block 1, m_{2} and v_{2} for block 2, and m_{f} and v_{f} for the resultant combined mass block, where m_{f} = 3m.

Total initial kinetic energy K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2

Since v_{2}=0, K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2

Total final kinetic energy K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2

Performing the same operation for momentum:

m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f}

Solving for v_{f} will yield v_{f} = \frac{2}{3}v_{1}

Plug this into the equation for final kinetic energy, and you get K_{f} = \frac{2}{3}mv_{1}^2

This implies that 1/3 of the energy is lost.
Alternate Solution - Unverified
thesandrus86
2011-04-24 14:21:41
por conservación del momento tenemos:
2mVi+mVi=(m+2m)Vf

2m =3mVf

Vf= 2/3....
NEC
archard
2010-04-11 14:54:59
Another quick way is to remember that kinetic energy can be written as \frac{p^2}{2m}, and since momentum is conserved, the final:initial kinetic energy ratio is \frac{2(m1)}{2(m1+m2)} = \frac{2}{3}. Which means the fractional kinetic energy loss is \frac{1}{3}NEC
wittensdog
2009-10-08 16:29:07
Here's something that's surprisingly useful. When a moving body hits a second one at rest, and they stick, the fractional loss of kinetic energy is,

m2 / ( m1 + m2)

If you just memorize that, all sorts of problems involving sticky collisions become way faster, even ones where you have falling masses on pendulums and so forth.
Dhananjay
2013-05-30 00:26:45
is m2 the mass of the moving ball or the ball at rest?
NEC
herrphysik
2006-09-19 16:14:15
Small typo: You're missing a 2 in the left side of the initial KE equation.Typo Alert!

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por conservación del momento tenemos: 2mVi+mVi=(m+2m)Vf 2m =3mVf Vf= 2/3....

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