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Nadusha1986 [10]
3 years ago
11

billiardballsnew2 A white billiard ball with mass mw = 1.49 kg is moving directly to the right with a speed of v = 3.09 m/s and

collides elastically with a black billiard ball with the same mass mb = 1.49 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 29° and the black ball ends up moving at an angle below the horizontal of θb = 61°. 1)What is the final speed of the white ball? m/s 2)What is the final speed of the black ball?
Physics
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

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Answer:

25.6 m/s

Explanation:

Draw a free body diagram of the sled.  There are two forces acting on the sled:

Normal force pushing perpendicular to the hill

Weight force pulling straight down

Take sum of the forces parallel to the hill:

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 38.0°)

a = 6.03 m/s²

Given:

v₀ = 0 m/s

a = 6.03 m/s²

t = 4.24 s

Find: v

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v = 25.6 m/s

5 0
2 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0
2 years ago
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