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Mars2501 [29]
3 years ago
14

A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s

. (a) Why does the tackle constitute a perfectly inelastic collision? (b) Calculate the velocity of the players immediately after the tackle and (c) determine the mechanical energy that is lost as a result of the collision. (d) Where did the lost energy go?
Physics
1 answer:
V125BC [204]3 years ago
7 0

Answer:

a) Please see below as the answer is self-explanatory.

b) 2.88 m/s

c) 785. 8 J

d) It is expended like thermal energy, due to internal friction.

Explanation:

a) In a tackle, both players keep emmeshed each other, so it is a perfectly inelastic collision; Immediately after the tackle, both masses behave like they were only one.

b) Assuming no external forces act during the collision, total momentum must be conserved.

As momentum is a vector, the conservation principle must be met by all vector components at the same time.

In our case, as the players move in directions mutually perpendicular, we can decompose the momentum vector along both directions, taking into account that after the collision, the momentum vector will have components along both directions.

So, if we call the W-E axis our X-axis (being the direction towards east as the positive one) , and to the S-N axis our Y -axis (being the northward direction the positive one), we can write the following equations:

pₓ₀ = pₓf ⇒ m₁*v₁ = (m₁+m₂)*vf*cosθ

py₀ = pyf ⇒ m₂*v₂ = (m₁+m₂)*vf*sin θ

where θ, is the angle that both players take regarding the x-axis after the collision (north of east).

Replacing by the values, we have the following equations:

vf*cosθ = (90.0 kg*5.00 m/s) / (90.0 kg + 95.0 kg) = 2.43 m/s (1)

vf*sin θ = (95.0 kg* 3.00 m/s) / (90.0 kg + 95.0 kg) = 1.54 m/s (2)

Dividing both sides:

sin θ / cos θ = tan θ = 1.54 / 2.43 = 0.634

⇒ arc tan (0.634) = 32.3º

Replacing in (1) we have:

vf = 2.43 m/s / cos 32.3º = 2.43 m/s / 0.845 = 2.88 m/s

c) As the collision happens in one dimension, all mechanical energy, before and after the collision, is just the kinetic energy of the players.

Before the collision:

K₀ = 1/2*m₁*v₁₀² + 1/2 m₂*v₂₀²

= 1/2*( ( 90.0) kg*(5.0)²(m/s)² + (95.0)kg*(3.0)(m/s)²) = 1,553 J

After the collision:

Kf = 1/2 *(m₁+ 767.2 Jm₂)*vf² = 1/2*185 kg*(2.88)²(m/s)²= 767.2 J

The mechanical energy lost during the collision is just the difference between the final and initial kinetic energy:

ΔK = Kf - K₀ = 767.2 - 1,553 J = -785.8 J

So, the magnitude of the energy lost during the collision is 785.8 J.

d) This energy is lost during the collision as thermal energy, due to the internal friction between both players.

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4 0
2 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

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#LearnwithBrainly

8 0
3 years ago
Si la fuerza de repulsión entre dos cargas es 18 × 1013
sattari [20]

Answer:

Explanation:

F = kQq/r²

r = √(kQq/F)

a)  r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m

  r = 40 mm

b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m

   r = 42 mm

5 0
3 years ago
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