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alexdok [17]
3 years ago
14

Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the othe

r is made of rubber. The steel bullet has a perfectly inelastic collision with the block, while the rubber bullet has an elastic collision. Which bullet is more likely to knock over the block, or are both equally likely to do so? Justify your choice based on physics principles.
Physics
1 answer:
tatiyna3 years ago
8 0

Answer and Explanation:

  • Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.
  • Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.
  • One contrast that will be effectively obvious is that the weapon in the case of heavy bullet will recoil more.  
  • One can consider energy assimilation as force times separation distance, and energy ingestion as a product of force and time.
  • Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
  • It will along these lines, additionally, give a more noteworthy "kick" to the object that absorbs.
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Answer:

Explanation:

According to a free body diagram the forces in the horizontal direction on body 1 would be:

F₁ = a₁*m₁ = -N

and on body 2:

F₂ = a₂*m₂ = N - F

N: normal force between the two blocks

F: frictional force on block 2

Since the two blocks are moving together, they need to have the same acceleration:

a₁ = a₂

This gives two equations with two unknown. Solving for a and N gives:

a = - \frac{F}{m_1+m_2}

N = -am_1

case A:

|a| = 0.96 m/s²

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case B:

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Answer:

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A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00
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Answer:

a) 200 m

b) 100 m/s

c) 709 m

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Explanation:

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Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

V0 = 0

Y(t) = 1/2 * 25 * t^2

Y(t) = 12.5 * t^2

And speed under constant acceleration:

Vy(t) = Vy0 + a * t

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It burns for 4 s and runs out of fuel

Y(4) = 12.5 * 4^2 = 200 m

V(4) = 25 * 4 = 100 m/s

Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity

It will be under constant acceleration. These new equations will have different starting constants.

Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2

Vy(t) = Vy4 + g * (t - 4)

When it reaches its highest point it will have a speed of zero.

0 = Vy4 + g * (t - 4)

0 = 100 - 9.81 * (t - 4)

100 = 9.81 * (t - 4)

t - 4 = 100 / 9.81

t = 10.2 + 4 = 14.2 s

At that moment it will have a height of:

Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m

The rocket will fall and hit the ground:

Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2

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t = 26.24 s

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Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s

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