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joja [24]
3 years ago
9

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.814 m/s at a 33.3° angle. What

is the x-component and y-component of the velocity of the second ball? pls help quick
Physics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

X - component = v2 = 0.650 m/s

Y - component = v2 = 0.447 m/s

Explanation:

Given that :

Conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

m = mass of pool ball ; u = initial velocity ; v = final velocity

Since the masses are identical

x - component

u1 + u2 = v1 + v2

u1 = 1.33 m/s

u2 = 0

Final speed of first ball = 0.814 m/s at an angle of 33.3° ;

v1 = 0.814*cos33.3 = 0.814 * 0.8358073 = 0.6803471 m/s

v2x = u1x + u2x - v1

v2x = 1.33 + 0 - 0.6803471

v2x = 0.6496529 m/s

v2x = 0.650 m/s

Y - component :

v1sinθ= 0.814*sin33.3 ; 0.814 * 0.5490228 = 0.4469045 m/s

v2y = 0.447 m/s

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(b) the kinetic energy of the bullet plus the block before the collision is 500J

(c) the kinetic energy of the bullet plus the block after the collision is 16.13J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki =  ¹/₂(0.1 x 100²) +  ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

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4 years ago
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