Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass m₂ = 39.0 k

Question

Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass m₂ = 39.0 kg, which was initially at rest. The blocks stick together after the collision.
A: Find the magnitude π of the total initial momentum of the two-block system. Express your answer numerically.
B: Find v_f, the magnitude of the final velocity of the two-block system. Express your answer numerically.
C: what is the change \Delta K= K_{final}- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

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Philomena 5 years 2021-08-29T09:47:30+00:00 1 Answers 6 views 0

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  1. Maris
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    2021-08-29T09:48:37+00:00
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    Answer:

    a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

    Explanation:

    a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

    The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

    So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

    b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

    p₂ = (1.3 + 39.0)v = 40.3v

    From the principle of conservation of momentum,

    p₁ = p₂

    37.7 kgm/s = 40.3v

    v = 37.7/40.3 = 0.94 m/s

    So the final velocity of the two-block system is 0.94 m/s

    c. The change in kinetic energy of the two-block system is ΔK = K₂ – K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

    So, ΔK = K₂ – K₁ = 1/2(m₁ + m₂)v² – 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² – 1/2 × 1.3 × 29.0² = 17.805 J – 546.65 J = -528.845 J ≅ -528.85 J

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