Difference between revisions of "Forest UCM MnAM InElasticCol"

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An Inelastic collision conservers Momentum But Not energy
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An Inelastic collision conserves Momentum But Not energy
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 +
:<math>\vec{P}_{\mbox{initial}} = \vec{P}_{\mbox{final}}</math>
 +
 
  
 
Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively.  They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.
 
Consider a collision between two bodies of mass <math>m_1</math> and <math>m_2</math> initially moving at speeds <math>v_1</math> and <math>v_2</math> respectively.  They stick together after they collide so that each is moving at the same velocity <math>v</math> after the collision.
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:<math>m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v</math>
 
:<math>m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v</math>
  
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Given that the amsses and initially velocities are known we can solve for v such that  
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Given that the masses and initially velocities are known we can solve for <math>v</math> such that  
  
 
:<math>v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>
 
:<math>v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}</math>
  
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[[Forest_UCM_MnAM#Inelastic_Collision_of_2_bodies]]
 
  
 
= problem 3.4 two hobos=
 
= problem 3.4 two hobos=
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;Conservation of momentum
 
;Conservation of momentum
 
:<math>0 = m(u-v_1) - (m+M) v_1</math>
 
:<math>0 = m(u-v_1) - (m+M) v_1</math>
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:<math>v_1 = \frac{2m}{M+2m}u</math>: same as before
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:<math>v_1 = \frac{m}{M+2m}u</math>: no factor of 2 like before
  
 
But now the second hobo jumps off
 
But now the second hobo jumps off
  
 
:<math>m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2</math>
 
:<math>m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2</math>
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: <math>-(m+M)\frac{2m}{M+2m}u=  mu- (M+m) v_2</math>
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: <math>-(m+M)\frac{m}{M+2m}u=  mu- (M+m) v_2</math>
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: <math>(M+m) v_2=  mu + (m+M)\frac{2m}{M+2m}u</math>
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: <math>(M+m) v_2=  mu + (m+M)\frac{m}{M+2m}u</math>
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: <math>(M+m) v_2=  \frac{m(M+2m)+2m(m+M)}{M+2m}u</math>
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: <math>(M+m) v_2=  \frac{m(M+2m)+m(m+M)}{M+2m}u</math>
 +
: <math>(M+m) v_2=  \frac{m(M+2m+m+M)}{M+2m}u</math>
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: <math>(M+m) v_2=  \frac{m(2M+3m)}{M+2m}u</math>
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: <math> v_2=  \frac{m(2M+3m)}{(M+2m)(M+m)}u</math>
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 +
if written in terms of the velocity from part (a)
 +
 
 +
: <math> v_2=  \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v</math>
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The flatcar will be moving faster if the hobos jump off separately.
 +
 
 +
[[Forest_UCM_MnAM#Inelastic_Collision_of_2_bodies]]

Latest revision as of 15:40, 15 September 2014

An Inelastic collision conserves Momentum But Not energy

[math]\vec{P}_{\mbox{initial}} = \vec{P}_{\mbox{final}}[/math]


Consider a collision between two bodies of mass [math]m_1[/math] and [math]m_2[/math] initially moving at speeds [math]v_1[/math] and [math]v_2[/math] respectively. They stick together after they collide so that each is moving at the same velocity [math]v[/math] after the collision.


If there are no external force then

Conservation of momentum
[math]m_1 v_1 + m_2 v_2 = \left (m_1 + m_2 \right ) v[/math]

Given that the masses and initially velocities are known we can solve for [math]v[/math] such that

[math]v= \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}[/math]


problem 3.4 two hobos

Two hobos are standing at one end of a stationary railroad flatcar of mass [math]M[/math]. Each hobo has a mass [math]m[/math] and the flatcar has frictionless wheels. A hobo can run to the other end of the car and jump off with a speed [math]u[/math] with respect to the car.

a.) Find the final speed of the car if both men run and jump off simultaneously.


At the instant the hobo jumps off with speed [math]u[/math] the railcar will move in the opposite direction at some speed [math]v[/math]. Since the hobo's speed is relative to the car, the hobos speed relative to the ground is [math]u-v[/math].

conservationof momentum
[math]\Rightarrow 0 = 2m(u-v) - Mv[/math]

or

[math]v = \frac{2m}{M+2m}u[/math]

b.) Now consider the case where they jump separately. The second hobo starts running AFTER the first hobo has jumped off.

Break the problem up into the two parts

PART A: The first hobo jumps off

Conservation of momentum
[math]0 = m(u-v_1) - (m+M) v_1[/math]
[math]v_1 = \frac{m}{M+2m}u[/math]: no factor of 2 like before

But now the second hobo jumps off

[math]m(u-v_1) - (m+M) v_1 = m(u-v_1) + m(u-v_2) - (M) v_2[/math]
[math]-(m+M)\frac{m}{M+2m}u= mu- (M+m) v_2[/math]
[math](M+m) v_2= mu + (m+M)\frac{m}{M+2m}u[/math]
[math](M+m) v_2= \frac{m(M+2m)+m(m+M)}{M+2m}u[/math]
[math](M+m) v_2= \frac{m(M+2m+m+M)}{M+2m}u[/math]
[math](M+m) v_2= \frac{m(2M+3m)}{M+2m}u[/math]
[math] v_2= \frac{m(2M+3m)}{(M+2m)(M+m)}u[/math]

if written in terms of the velocity from part (a)

[math] v_2= \frac{2M+3m}{2M+2m}\frac{2m}{M+2m}u =\frac{2M+3m}{2M+2m}v[/math]

The flatcar will be moving faster if the hobos jump off separately.

Forest_UCM_MnAM#Inelastic_Collision_of_2_bodies

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