The total momentum of two marbles before collision is 0.06 kg-m/s. no outside forces act on the marbles. what is the total momentum of the marbles after the collision

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



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

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.