Respuesta :
if you want individual total momentum,
it depends
- on the masses of the marbles
- on the velocities before impact
- on the velocities after impact
- was it elastic pre/post impact (bounced off)?
- was it inelastic pre/post impact (stuck a)?
with the given info, we must assume masses and velocities of the two marbles are the same which means there are 4 possible answers. If the question is suggesting that their INDIVIDUAL momentums were 0.12 kg*m/s vs. their COMBINED momentum (0.06 kg*m/s) before impact.
1. combined momentum & inelastic
0.06 kg*m/s = (m+m)*v
0.03 kg*m/s = m×v
2. individual momentum & inelastic
0.12 kg*m/s = (m+m)*v
0.06 kg*m/s = m×v
3. combined momentum & elastic
0.06 kg*m/s = m×v1f + m×v2f (b/c vf is different)
0.06 kg*m/s = m×(v1f + v2f)
4. individual momentum & elastic
0.12 kg*m/s = m×v1f + m×v2f (b/c vf is same)
0.06 kg*m/s = m×v
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
IF you want COMBINED total momentum, apply law of conservation of momentum which states that momentum is conserved without external forces... so it would be the same.
0.06 kg×m/s
it depends
- on the masses of the marbles
- on the velocities before impact
- on the velocities after impact
- was it elastic pre/post impact (bounced off)?
- was it inelastic pre/post impact (stuck a)?
with the given info, we must assume masses and velocities of the two marbles are the same which means there are 4 possible answers. If the question is suggesting that their INDIVIDUAL momentums were 0.12 kg*m/s vs. their COMBINED momentum (0.06 kg*m/s) before impact.
1. combined momentum & inelastic
0.06 kg*m/s = (m+m)*v
0.03 kg*m/s = m×v
2. individual momentum & inelastic
0.12 kg*m/s = (m+m)*v
0.06 kg*m/s = m×v
3. combined momentum & elastic
0.06 kg*m/s = m×v1f + m×v2f (b/c vf is different)
0.06 kg*m/s = m×(v1f + v2f)
4. individual momentum & elastic
0.12 kg*m/s = m×v1f + m×v2f (b/c vf is same)
0.06 kg*m/s = m×v
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
IF you want COMBINED total momentum, apply law of conservation of momentum which states that momentum is conserved without external forces... so it would be the same.
0.06 kg×m/s
The total momentum of two marbles before collision is 0.06 kg-m/s. The total momentum of the marbles after the collision is 0.06 kg-m/s.
What is conservation of momentum principle?
When there is no external force acting on the system, the total momentum of the system remains conserved.
In this situation, the momentum of the system will remain conserved as no outside force acts on the marbles.
Initial momentum i.e. before collision = final momentum (after collision)
0.06kg-m/s =0.06kg-m/s
Thus, the total momentum of the marbles after the collision is 0.06 kg-m/s.
Learn more about conservation of momentum principle.
https://brainly.com/question/14033058
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