Dee asked in Science & MathematicsPhysics · 3 weeks ago

A 100g block on a frictionless table is firmly attached to one end of a spring with k = 20N/m. ?

A 100g block on a frictionless table is firmly attached to one end of a spring with k = 20N/m. The other end of the spring is anchored to the wall. A 20g ball is thrown horizontally toward the block with a speed of 5.0m/s.

a) If the collision is perfectly elastic, what is the balls speed immediately after the collision?

b) What is the maximum compression of the spring?

c) Repeat parts a and b for the case o perfectly inelastic collision 

1 Answer

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  • NCS
    Lv 7
    3 weeks ago
    Favorite Answer

    a) For an elastic, head-on collision, we know (from conservation of energy), that the relative velocity of approach = relative velocity of separation, or

    5.0 m/s = u - v

    where v is the post-collision velocity of the ball

    and u is the post-collision velocity of the block

    so

    u = v + 5.0m/s

    conserve momentum for the collision, substituting for u:

    20g*5.0m/s = 20g*v + 100g*(v + 5.0m/s)

    solves to

    v = -3.33 m/s

    and so the ball's speed is 3.33 m/s ◄

    b) Then u = 1.67 m/s, and the block's KE gets converted into spring PE:

    ½ * 0.100kg * (1.67m/s)² = ½ * 20N/m * x²

    solves to

    x = 0.12 m ◄

    c) Now the objects have a common post-collision velocity:

    20g*5.0m/s = 120g*v

    solves to

    v = 0.83 m/s ◄

    ½*0.120kg*(0.83m/s)² = ½ * 20N/m * x²

    solves to

    x = 0.065 m ◄

    Hope this helps!

    Source(s): Hey, was this working helpful? https://answers.yahoo.com/question/index?qid=20191...
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