GREPhysics.NET: GR9677 #7

GR9677 #7

Problem

GREPhysics.NET Official Solution

Alternate Solutions

This problem is still being typed.

Mechanics $\Rightarrow$ }Elastic Collisions

One determines the velocity of impact of the ball from conservation of energy,

$mgh = \frac{1}{2}mv_0^2 \Rightarrow v_0^2 = 2 gh $

Conservation of momentum gives,

$v_0=-v_1 + 2v_2 $

Conservation of kinetic energy gives,

$mv_0^2 = mv_1^2+2v_2^2 $

Plug in the momentum and kinetic energy conservation equations to solve for $v_1$ and $v_2$ in terms of $v_0$ to get

$\begin{eqnarray} v_1&=&-v_0/3\\ v_2&=2v_0/3& \end{eqnarray}$

Write yet another conservation of energy equation for the final energy,

$\frac{1}{2}mv_1^2 +\frac{1}{2}2mv_2^2 = mgh' +\frac{1}{2}2m v_2^2, $

where the condition that the mass $2m$ slides on a frictionless plane is used.

Thus, $\frac{1}{2}mv_1^2 = \frac{v0^2}{18} = mgh'\Rightarrow h'=\frac{h}{9}$ , where the previous result $v_1 = -v_0/3$ and $v_0^2 =2gh$ is used.

Alternate Solutions

nbwell
2007-10-31 12:08:48

There is an easier way using conservation of momentum and energy. Consider the ratio of the masses m1 (the ball) and m2 (the brick). The initial momentum into the system is going to be m1 times the initial velocity of m1. After the collision, the momentum is divided between the masses so that the momentum (magnitude) of m1 is half the momentum (magnitude) of m2; put another way, the post collision momentum (magnitude) of m1 is 1/3 of the total initial momentum. Thus we've conserved momentum.

Now to determine the height consider conservation of energy. Momentum is proportional to velocity, and kinetic energy is proportional to velocity squared. Since post collision m1 has 1/3 its initial momentum, it only has 1/9 its initial kinetic energy. As it flies upwards, the kinetic energy is converted to potential energy (mgh). So with only 1/9 the kinetic energy, it can only achieve 1/9 the potential energy and thus only rise to 1/9 its initial height. (you could also have realized this since kinetic energy is proportional to momentum squared)

In short, inspect the ratio of masses to determine how the momentum is going to be divided. Deduce how the reduced momentum will reduce your kinetic energy. Finally realize that this reduction in kinetic energy will equal the reduction in potential energy and the reduction in height.

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Comments

nbwell
2007-10-31 12:08:48

blah22
2008-02-14 09:48:43

I disagree with this method. I think your reasoning only works for the specific case of mass ratio of 1:2.

If the ratio were 1:3, by your reasoning after the collision m1 should have 1/3 of m2's momentum so m1 will have 1/4 of the total initial momentum, correct? Then you would square this number and the final height would only be 1/16 of its initial height. But if you run the correct formula that Richard posted below, you see that after such a collision m1's velocity is half of its initial velocity after collision, squaring THAT you get a final height of 1/4 initial height.

The latter solution seems more plausible to me, as you increase the relative mass of m2, m1 should gain more height...not less. Going from 1/9 -> 1/4 rather than 1/9 -> 1/16 with your reasoning. Because if you toss a ball at the earth it returns with nearly all of its energy.

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Richard
2007-09-24 23:45:12

As much as I like all of your solutions, I don't think they are really viable in 1.7 minutes. This is how I think it should be done:
You should recall that when you have a perfectly elastic collision with one mass initially at rest, there is a very succint relation between the velocites, namely,
$v_{1,f}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1,i}.$
$v_{2,f}=\frac{2m_{1}}{m_{1}+m_{2}}v_{1,i}.$
Of course we only need the first one: $v_{1,f}=1/3v_{1,i}$ .
Then, realizing (remembering) that $v_{i}^2=2gh_{i}$ , and using conservation of energy, $1/2mv^2=mgh_{f}$ we have,
$gh_{i}/9=gh_{f}\rightarrow h_{f}=h_{i}/9$
Life is so much easier when you remember the right things.

phys2718
2008-10-03 11:38:24

Also note the similarity of the relation between the initial and final velocities to the relation of the transmission and reflection coefficients for waves crossing from one medium to another. I think Richard is right, these expressions are just worth memorizing for a number of different problems.

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cyberdeathreaper
2006-12-31 15:35:53

The direction of v1 is indeed negative, but its scalar value is not. Therefore, this equation is correct:

$v_0 = -v_1 + 2v_2$

but this solution is not:

$v_1 = -(1/3)v_0$

If it were, then your assumption that v1 moves to the left would be wrong. The correct answer is:

$v_1 = (1/3)v_0$

(See my algebra in the other post to verify how I got this value)

NOTE: The reason the typo doesn't impact the final answer is because v_1 is squared in the final equation, eliminating its sign.

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cyberdeathreaper
2006-12-30 16:23:08

For those interested, here is an expanded explanation of how to get v1 and v2:

First solve for v1 in terms of v0 and v2 from your conservation of momentum equation:

$v_0 = -v_1 + 2v_2$

$v_1 = 2v_2 - v_0$

Now solve for v2 in terms of v0 from your conservation of kinetic energy equation, substituting for v1 the equation we just found above:

$v^2_0 = v^2_1 + 2v^2_2$

$v^2_0 = (2v_2 - v_0)^2 + 2v^2_2$

$v^2_0 = 4v^2_2 - 4(v_0)(v_2) + v^2_0 + 2v^2_2$

$v^2_0 = v_2(6v_2 - 4v_0) + v^2_0$

$0 = v_2(6v_2 - 4v_0)$

$6v_2 = 4v_0$

$v_2 = (2/3)v_0$

Now substitute v2 back into your equation for v1, to get:

$v_1 = 2v_2 - v_0$

$v_1 = (4/3)v_0 - v_0$

$v_1 = (1/3)v_0$

Thus giving you the answers obtained in the solution provided.

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senatez
2006-11-01 10:44:16

Can you explain in greater detail? I can solve it, but I run into some messy algebra with quadratic equations. Can you explain how you so elegantly obtained these values?

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yubs
2006-10-22 18:25:19

(Sorry, it won't let me edit my previous post, tells me I should login, but I am logged in...)

I'm a bit confused here and I think there may be a typo.

In the third equation, is a 'm' missing from the third term? Shouldn't it read:

$mv_0^2 = mv_1^2+2mv_2^2$

Also, I'm having trouble getting your solution for $v_1$ & $v_2$ . Could you elaborate on "plug in the equations?

many thanks!

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yubs
2006-10-22 18:10:31

I'm a bit confused here and I think there may be some typos.

I think there is a 'm' missing from the third equation. Shouldn't it read:

mV_o^2 = mV_1^2 + 2mV_2^2

Also, I'm having trouble getting your solution for V_1 & V_2. Could you elaborate on "plug in the equations?

many thanks!

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