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geniusboy [140]
2 years ago
7

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

stationary and rebounds with 70% of its initial kinetic energy. What is the magnitude of the change in momentum of object A?
Physics
1 answer:
AnnZ [28]2 years ago
6 0

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

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Answer: 42.49 m/s^{2}

Explanation:

To solve this, we need to keep in mind the following:

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Tangent (∅) = Opossite Side / Adyacent Side

By trigonometry, we know the previous formula. This formula allows us to find the Tangent of a angle as a relation between the two perpendiculars magnitudes. In this case the Opossite Side will be the Gravity Accelaration, while the Adyancent Side is the Van's Acceleration.

(1)  Tangent (∅) = Gravity's Acceleration (G) / Van's Acceleration (Va)        

Searching for the Va in (1)

Va = G/Tan(∅)

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Va = 9.81m/s^{2}  / Tan(13.0°)

Va = 42.49 m/s^{2}

The vans acceleration need to be 42.49 m/s^{2}  to create an angle of 13° with the Van's Roof

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The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
 We have then:
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 Then, the resulting net force is:
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 answer:
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 2.941756275 m / s ^ 2
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