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2 years ago

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Two large blocks of wood are sliding toward each other on the frictionless surface of a frozen pond. Block A has mass 4.00 kg an

d is initially sliding east at 2.00 m/s. Block B has mass 6.00 kg and is initially sliding west at 2.50 m/s. The blocks collide head-on. After the collision block B is sliding east at 0.50 m/s. What is the decrease in the total kinetic energy of the two blocks as a result of the collision? Express your answer with the appropriate units.

1 answer:

ollegr [7]2 years ago

5 0

Answer:

13.50 J

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, so we can write the following equation:

Δp = 0 ⇒ p₁ = p₀

Assuming that the mass moving to the east has positive speed, we can write:

p₀ (initial momentum) = 4.00 kg*2.00 m/s + 6.00kg*(-2,5 m/s) = -7.00 kg*m/s

p₁ (final momentum) = 4.00kg*vAm/s + 6.00kg*(0.5m/s) = -7.00 kg*m/s

Solving for vA, we have:

vA = -2.50 m/s

Now, we can find the initial and final kinetic energies, as follows:

Ki = 1/2*4.00kg*(2.00)²(m/s)² +1/2*6.00*(-2.50)²(m/s)² = 26.75 J

Kf= 1/2*4.00kg*(-2.50)²(m/s)² +1/2*6.00*(0.50)²(m/s)² = 13.25 J

⇒ ΔK = Kf-Ki = 13.25 J - 26.75 J = -13.50 J

So, the decrease in the total kinetic energy of the two blocks as a result of the collision is equal to 13.50 J.

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Answer:

a) a = g / 3

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Explanation:

Given:-

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$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

Where, a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0)

m (3.0) = 2.00*14.7

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Answer:

The ball slows down in the air due to an unbalanced force

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When player kicks the ball, there are mainly two foces acting on this object: the force made by the player and the opposite force of gravity (which acts with a direction always to the centre of the Earth)

The force applied by the player will be decreasing, while the force of gravity is always constant, this will make that both forces will unbalance, making the football´s speed slow down

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Answer:

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Lets take amplitude = A

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• the net <u>vertical</u> force on the <u>larger</u> block is

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• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

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(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

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