Yahoo Answers: Answers and Comments for A 100g block on a frictionless table is firmly attached to one end of a spring with k = 20N/m. ? [Physics]
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Thu, 14 Nov 2019 08:00:52 +0000
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Yahoo Answers: Answers and Comments for A 100g block on a frictionless table is firmly attached to one end of a spring with k = 20N/m. ? [Physics]
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https://answers.yahoo.com/question/index?qid=20191114080052AA1afrr
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From NCS: a) For an elastic, headon collision, we know ...
https://answers.yahoo.com/question/index?qid=20191114080052AA1afrr
https://answers.yahoo.com/question/index?qid=20191114080052AA1afrr
Thu, 14 Nov 2019 12:46:29 +0000
a) For an elastic, headon 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 postcollision velocity of the ball
and u is the postcollision 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 postcollision 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!