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A bullet of mass 0.107 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3 kg that is sitting at rest on a nearly frictionless surface. What was the transfer of energy (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that represents energy transfer due to a temperature difference between a system and its surroundings.)

Respuesta :

aabdulsamir

Answer:

165.77J

Explanation:

M₁ = 0.107kg

u₁ = 300m/s

m₂ = 3kg

u₂ = 0

v =

m₁u₁ + m₂u₂ = (m₁ + m₂)V

(0.107*300) + 0 = (0.107 + 3)V

V = 32.1 / 3.107 = 10.33m/s

kinetic energy of the system after collision =

½m1v² + ½m2v²

K.E = ½(m1 + m2)v²

K.E = ½(0.107+3) * 10.33²

K.E = 165.77J

Cricetus

There will be no energy transferred to surroundings. To understand the calculation, check below.

Energy conservation

According to the question,

Bullet's Mass, m = 0.107 kg

Speed, v = 300 m/s

Block's Mass, = 3 kg

By using the Energy conservation,

⇒ [tex]\Delta E_{Thermal, Bullet}[/tex] + [tex]\Delta E_{Thermal, Block}[/tex] = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] mv²

By substituting the values, we get

                                                            = [tex]\frac{1}{2}[/tex] × 0.107 × (300)² + 0

                                                            = [tex]\frac{1}{2}[/tex] × 0.107 × 900

                                                            = 4815 J  

Thus the answer above is appropriate.

Find out more information about energy conservation here:

https://brainly.com/question/14274074

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