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tatyana61 [14]
3 years ago
15

In a closed system with no friction, a red sphere of 2.5 kg stands stationary. A blue sphere with a mass of 5.8 kg approaches th

e first sphere with a speed of 4.1 m/s. The two collide. After the collision, the blue sphere begins moving forward with a speed of 1.3 m/s. What is the velocity of the red sphere after the collision? (Show all work)
Physics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

m1 = the mass of the blue sphere = 5.8 kg

m2 = the mass of the red sphere = 2.5 kg

v1 = initial velocity of the blue sphere before the collision = 4.1 m/s

v2 = initial velocity of the red sphere before the collision = 0 m/s

v'1 = final velocity of the blue sphere after the collision = 1.3 m/s

v'2 = final velocity of the red sphere after the collision = ?

using conservation of momentum

m1v1 + m2v2 = m1v'1 + m2v'2

(5.8) (4.1) + (2.5) (0) = (5.8) (1.3) + (2.5) (v'2)

23.78 = 7.54 + (2.5) (v'2)

-7.54    -7.54

16.24 = (2.5) v'2

---------   -----------

 2.5         2.5

v'2 = 6.5

Explanation:

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Answer:

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Also, r=l\ sin\theta

Equation (1) becomes :

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Let t is the time taken for the ball to rotate once around the axis. It is given by :

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T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

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On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

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3 years ago
If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?
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More force

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Object A has more mass than object B

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Answer:

1) v₀x = 13.76 m/s

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Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

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then

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Solving this equation we get

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The distance between the two girls is 55.1746 m

5) and 6) If   v₀x = 15 m/s = vx   and   ymax = 24 m

y = ?   when x = (xmax/2)

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α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

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Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

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8 0
3 years ago
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