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Slav-nsk [51]
1 year ago
12

Cart 1 has an initial velocity and hits cart 2 which is stationary. after a perfectly inelastic collision, the combined carts ar

e traveling with exactly one third the initial velocity of cart 1, and in the same direction. which is correct about the mass of cart 2? the mass of cart 2 must be twice the mass of cart 1. the mass of cart 2 must have decreased during the collision. the mass of cart 2 must be one third the mass of cart 1. the mass of cart 2 must be the same as the mass of cart 1.
Physics
1 answer:
Tomtit [17]1 year ago
4 0

Option(a)  the mass of cart 2 is twice that of the mass of cart 1 is the right answer.

The mass of cart 2 is twice that of the mass of cart 1  is correct about the mass of cart 2.

Let's demonstrate the issue using variables:

Let,

m1=mass of cart 1

m2=mass of cart 2

v1 = velocity of cart 1 before collision

v2 = velocity of cart 2 before collision

v' = velocity of the carts after collision

Using the conservation of momentum for perfectly inelastic collisions:

m1v1 + m2v2 = (m1 + m2)v'

v2 = 0 because it is stationary

v' = 1/3*v1

m1v1 = (m1+m2)(1/3)(v1)

m1 = 1/3*m1 + 1/3*m2

1/3*m2 = m1 - 1/3*m1

1/3*m2 = 2/3*m1

m2 = 2m1

From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.

To learn more about inelastic collision visit:

brainly.com/question/14521843

#SPJ4

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Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

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An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

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Here, u = - 100 cm

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