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Sphere A of mass 400 g is travelling horizontally with a speed of 6 m s⁻¹ when it collides with sphere B of mass 150 g travelling in the opposite direction with a speed of 9 m s⁻¹ - Leaving Cert Physics - Question b - 2017

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Sphere A of mass 400 g is travelling horizontally with a speed of 6 m s⁻¹ when it collides with sphere B of mass 150 g travelling in the opposite direction with a sp... show full transcript

Worked Solution & Example Answer:Sphere A of mass 400 g is travelling horizontally with a speed of 6 m s⁻¹ when it collides with sphere B of mass 150 g travelling in the opposite direction with a speed of 9 m s⁻¹ - Leaving Cert Physics - Question b - 2017

Step 1

Apply the Law of Conservation of Momentum

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Answer

Using the conservation of momentum, we have the equation:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

  • m1m_1 = mass of sphere A = 0.4 kg (400 g)
  • u1u_1 = initial velocity of sphere A = 6 m s⁻¹
  • m2m_2 = mass of sphere B = 0.15 kg (150 g)
  • u2u_2 = initial velocity of sphere B = -9 m s⁻¹ (negative due to opposite direction)
  • v1v_1 = final velocity of sphere A = 0 m s⁻¹ (comes to rest)
  • v2v_2 = final velocity of sphere B (to be calculated)

Substituting the known values:

(0.4)(6)+(0.15)(9)=(0.4)(0)+(0.15)v2(0.4)(6) + (0.15)(-9) = (0.4)(0) + (0.15)v_2

Step 2

Solve for the New Velocity of Sphere B

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Answer

Rearranging the equation gives us:

2.41.35=0.15v22.4 - 1.35 = 0.15 v_2

Thus,

1.05=0.15v21.05 = 0.15 v_2

Now, divide both sides by 0.15:

v2=1.050.15=7ms1v_2 = \frac{1.05}{0.15} = 7 m s^{-1}

So, the new velocity of sphere B is 7 m s⁻¹.

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