Question

Let two objects of equal mass m collide. Object 1 has initial velocityv, directed to the right, and object 2 isinitially stationary.


If the collision is perfectly elastic, whatare the final velocities v_1 and v_2 of objects 1 and 2?

Answer 1

can show me with the actual equation--would be greatly appreicated

Answer 2

why V2=-v
i think V2= v

Answer 3

ok. My source for the language is wikipedia in dutch http://nl.wikipedia.org/wiki/Botsing_(natuurkunde)#Volkomen_elastische_botsing
For totally elasticical collisions there are two laws valid:
conservation of linear momentum (in one dimension case)
and conservation of kinetic energy
These laws were first deduced by Descartes, and ameliorated by Christian Huygens, while they were considering center of mass.

\[( m_1 v_1 + m_2 v_2)_{before coll.}= ( m_1 v_1 + m_2 v_2)_{after coll.}\]
and resp.
\[( \frac{m_1 {v_1^2}}2 + \frac{m_2 {v_2^2}}2)_{before coll.}= ( \frac{m_1 {v_1^2}}2 + \frac{m_2 {v_2^2}}2)_{after coll.}\]

from these two you can deduce the next equation:
\[v_{1after}-v_{2after}= -(v_{1before}-v_{2before})\]

Answer 4

@mos1635 you are right

Answer 5

correction:

v_1 = 0
v_2 = v

Answer 6

we have to use kinetic energy and conservation of momentum to figure out what the vs are? still confused a little

Answer 7

nath
aplay conservation of momentum
aplay conservation of kinetic energy
you now have 2 equations
solve system of equetions to V1 and V2...

Answer 8

m*v=M*V1+m*V2
v=V1+V2
v-V1=V2 (that is eq 1)

1/2mv^2=1/2mV1^2+1/2 m V2^2
V^2=V1^2+V2^2
V^2-V1^2=V2^2
(V-V1)*(V+V1)=V2^2
divide that with (1) and you have.....
V+V1=V2 (that is eq 2)

we now haveQ
v-V1=V2 and
V+V1=V2
sum those two and
2*V=2*V2=>V2=V
and by substitution to any of them V1=0