A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s.
What is the magnitude of the final velocity of the red marble?

A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s. 31 cm/s is the magnitude of the final velocity of the red marble. Solution 1.2 g × v_f = 56.7 g·cm/s - 19.25 g·cm/s v_f = (56.7 g·cm/s - 19.25 g·cm/s) / 1.2 g v_f = 31 cm/s

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Asked 13 days ago|7/5/2024 10:51:17 PM

Updated 8 days ago|7/10/2024 1:33:54 PM

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A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s. 31 cm/s is the magnitude of the final velocity of the red marble.

Solution

1.2 g × v_f = 56.7 g·cm/s - 19.25 g·cm/s

v_f = (56.7 g·cm/s - 19.25 g·cm/s) / 1.2 g

v_f = 31 cm/s

Solution

1.2 g × v_f = 56.7 g·cm/s - 19.25 g·cm/s

v_f = (56.7 g·cm/s - 19.25 g·cm/s) / 1.2 g

v_f = 31 cm/s

Added 13 days ago|7/5/2024 11:09:53 PM

This answer has been confirmed as correct and helpful.

1

m1=3.5 g=0.0035 kg

2=1.2g=0.0012 kg

m2=1.2 g=0.0012 kg

1 =15cm/s=0.15 m/s

v1i=15 cm/s=0.15 m/s

1 =5.5 cm/s=0.055m/s

v1f=5.5 cm/s=0.055 m/s

2i=3.5 cm/s=0.035 m/s

v2i=3.5 cm/s=0.035 m/s

Now, apply conservation of momentum:

1 1 + 2 2 = 1 1f+ 2 2

Substitute the known values:

0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012× 2

0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012×v2f

Calculate the left-hand side:

0.000525+0.000042=0.0001925+0.0012 2

0.000525+0.000042=0.0001925+0.0012v 2f

Simplify and solve for v2f:

0.0003745=0.0012v2f

v2f =0.0003745/0.0012

2 0.312 m/s

2=1.2g=0.0012 kg

m2=1.2 g=0.0012 kg

1 =15cm/s=0.15 m/s

v1i=15 cm/s=0.15 m/s

1 =5.5 cm/s=0.055m/s

v1f=5.5 cm/s=0.055 m/s

2i=3.5 cm/s=0.035 m/s

v2i=3.5 cm/s=0.035 m/s

Now, apply conservation of momentum:

1 1 + 2 2 = 1 1f+ 2 2

Substitute the known values:

0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012× 2

0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012×v2f

Calculate the left-hand side:

0.000525+0.000042=0.0001925+0.0012 2

0.000525+0.000042=0.0001925+0.0012v 2f

Simplify and solve for v2f:

0.0003745=0.0012v2f

v2f =0.0003745/0.0012

2 0.312 m/s

Added 8 days ago|7/10/2024 1:33:54 PM

Deleted by kanand [7/10/2024 1:34:12 PM], Undeleted by kanand [7/10/2024 1:34:15 PM], Rated good by kanand

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