GREPhysics.NET: GR0177 #4

GR0177 #4

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }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).

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 Reply to this comment
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. Reply to this comment

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.

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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

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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...

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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.

Reply to this comment

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.

Reply to this comment

thesandrus86
2011-04-24 14:21:41

por conservación del momento tenemos:
2mVi+mVi=(m+2m)Vf

2m =3mVf

Vf= 2/3....

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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 $\frac{1}{3}$

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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?

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herrphysik
2006-09-19 16:14:15

Small typo: You're missing a 2 in the left side of the initial KE equation.

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