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DIA [1.3K]
3 years ago
12

A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp

eed of 150 m/s and becomes embedded in the block. (a) Find the amplitude of the resulting SHM. (b) What percentage of the origi- nal kinetic energy of the bullet is transferred to mechanical energy of the oscillator
Physics
1 answer:
Yanka [14]3 years ago
6 0

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150  m/s , k = 500 N / m

a.

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v²  -  ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)²   /   (500 N / m)

A = 0.1656 m

b.

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

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Let velocity along the x-axis = Vx

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Applying law of conservation of momentum along x-axis

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9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

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b) Aplying conservation of momentum along the x-axis

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Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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