I need help finding the velocity of a 21.6 g marble after it collides. The scenario is as follows:

Question

21.6 g marble sliding to the right at 54.8 cm/s overtakes and collides elastically with a 10.8 g marble moving in the same direction at 13.7 cm/s. After the collision, the 10.8 g marble moves to the right at 68.5 cm/s. Find the velocity of the 21.6 g marble after the collision.

Answer 1

Convert to SI units (a good thing to do, although sometimes unnecessary, will save you grief more often than not)

21.6 g = 0.0216 kg

54.8 cm/s = 0.548 m/s

10.8 g = 0.0108 kg

13.7 cm/s = 0.137 m/s

68.5 cm/s = 0.685 m/s

Assume "right" is positive.

Let m_a = mass of the 21.6 g marble and m_b = mass of the 10.8 g marble

Let v_a-1 and v_a-2 = the velocity before and after the collision (respectively) of the 21.6 g marble

Let v_b-1 and v_b-2 = the velocity before and after the collision (respectively) of the 10.8 g marble

Before the collision:

Momentum: P₁ = m_av_a-1 + m_bv_b-1 = (0.0216)(0.548) + (0.0108)(0.137) = 0.133164 kg•m/s

Kinetic Energy E₁ = (1/2)m_av_a-1² + (1/2)m_bv_b-1² = (1/2)(0.0216)(0.548)² + (1/2)(0.0108)(0.137)² = 0.0033446 j

After the collision:

Momentum: P₂ = m_av_a-2 + m_bv_b-2 = (0.0216)v_a-2 + (0.0108)(0.685) = 0.0216v_a-2 + 0.007398

Kinetic Energy E₂ = (1/2)m_av_a-2² + (1/2)m_bv_b-2² = (1/2)(0.0216)(v_a-2)² + (1/2)(0.0108)(0.685)² = 0.0108v_a-2² + 0.002534

Conservation of momentum:

P₁ = P₂ so:

0.133164 = 0.0216v_a-2 + 0.007398

v_a-2 = 0.274

Conservation of Kinetic Energy:

E₁ = E₂ so:

0.0033446 = 0.0108v_a-2² + 0.002534

v_a-2 = 0.274

So the velocity of the 21.6 g marble is 0.274 m/s or 27.4 cm/s

We only needed to use EITHER the conservation of Momentum OR the Conservation of Kinetic Energy. I used both so you could see that the answer turns out the same in elastic collisions.

Answer 2

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'; v₂^' = (21.6g·54.8cm/s + 10.8g·13.7cm/s - 10.8g·68.5cm/s)/21.6g = 27.4 cm/s

in the same direction