A bowling ball of mass mb = 4.4 kg is rolled down the lane with a velocity of v1 = 2.3 m/s. It strikes a single remaining pin of mass mp = 0.82 kg head on. After the collision the bowling ball has a velocity (in the same direction) of v2 = 1.65 m/s. Write an expression for the velocity of the pin in the horizontal direction after the impact, vp. What is the pin's speed in m/s? Assuming the coefficient of friction is μk = 0.1, how far in meters could the pin slide before stopping if it doesn't topple?

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Description: A bowling ball of mass mb = 4.4 kg is rolled down the lane with a velocity of v1 = 2.3 m/s. It strikes a single remaining pin of mass mp = 0.82 kg head on. After the collision the bowling ball has a velocity (in the same direction) of v2 = 1.65 m/s.

The linear momentum of any object having mass m and velocity v is defined as, P→ = mv→ - - - (1)…

Answered: A bowling ball of mass mb = 4.4 kg is…

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A bowling ball of mass m_b = 4.4 kg is rolled down the lane with a velocity of v₁ = 2.3 m/s. It strikes a single remaining pin of mass m_p = 0.82 kg head on. After the collision the bowling ball has a velocity (in the same direction) of v₂ = 1.65 m/s.

Write an expression for the velocity of the pin in the horizontal direction after the impact, v_p.

What is the pin's speed in m/s?

Assuming the coefficient of friction is μ_k = 0.1, how far in meters could the pin slide before stopping if it doesn't topple?

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