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Elastic collision, one dimension

1. Homework Statement

bloks b and c have m asses 2m and m respecti vely, and are at rest on a firctionless surface

black a also of mass m.. is heading at speed v toward block b as show... determine te final velocity of each block after alll subsequent collisions are over, assum all collision are elastic

aaaaaaa.jpg


2. Homework Equations


v_1f = (m_1-m_2/(m_1+m_2)) * v_1i + (2m_2/(m_1+m_2)) * v_2i

and

v_2f = (2m_1/(m_1+m_2)) * v_1i + (m_2-m_1/(m_1+m_2)) * v_2i

3. The Attempt at a Solution

i compared block A when it hits block B and then compared block C when block B hits it..

i got...

(m-2m / 2m+ m) * v

block a) - 1/3 v

found block b's v_i before it hits block c to be 2/3 v

(2m - m / 3m) *2/3v

block b) 2/9 v

(2*2m / 3m) *2/3v

block c) 8/9v






i feel confident these answers are right but just trying to see what others think
 
Last edited:
454
0
1. Homework Statement

bloks b and c have m asses 2m and m respecti vely, and are at rest on a firctionless surface

black a also of mass m.. is heading at speed v toward block b as show... determine te final velocity of each block after alll subsequent collisions are over, assum all collision are elastic

aaaaaaa.jpg


2. Homework Equations


v_1f = (m_1-m_2/(m_1+m_2)) * v_1i + (2m_2/(m_1+m_2)) * v_2i

and

v_2f = (2m_1/(m_1+m_2)) * v_1i + (m_2-m_1/(m_1+m_2)) * v_2i

3. The Attempt at a Solution

i compared block A when it hits block B and then compared block C when block B hits it..

i got...

block a) - 1/3 v

found block b's v_i before it hits block c to be 1/3 v
Before a and b collide their total momentum is m_a * v

After the collision of a and b if v_a = -1/3 and v_b = 1/3

m_a*v_a + m_b*v_b = 1/3 m_a * v so momentum wasn't conserved
 
mv + mv = mv + mv
ai -- bi -- af --bf
1 + 0 = 1/3 + 2/3

1 = 1 = momentum conservation...

negative for direction
 
Last edited:
anyone verify my answer?
 
454
0
mv + mv = mv + mv
ai -- bi -- af --bf
1 + 0 = 1/3 + 2/3

1 = 1 = momentum conservation...

negative for direction
you had -1/3v for the speed of a. -1/3 + 2/3 isn't equal to 1.
If you substitute the right values in the formula for v2_f that you gave, you should get
the right value for the speed of b after the first collision.
 
v_2f = (2m_1/(m_1+m_2)) * v_1i + (m_2-m_1/(m_1+m_2)) * v_2i

v_2f = (2*m / 3m) * v + 0

v_2f = 2/3v
 
i think my answers now are correct, and the negitive on the 1/3 is for direction, dosnt mean that momentum is not being conserved
 
Last edited:

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