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Rasek [7]
2 years ago
10

Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one

is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Physics
1 answer:
Dvinal [7]2 years ago
6 0

Answer:

e.

Explanation:

  • Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according  Newton's 2nd Law, as we can see below:
  • F_{g}  = m*g = m*a  (1)
  • ⇒a = g = 9.8m/s² (pointing downward)
  • Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:

       \Delta y = v_{o} * t - \frac{1}{2} *g*t^{2}  (2)

  • Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:

        h =\frac{1}{2} *g*t^{2}  (3)

  • Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.
  • If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.
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A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re
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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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Answer:

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