A massless spring with spring constant k is compressed by an amount xo. A block

Question

A massless spring with spring constant k is compressed by an amount xo. A block of mass m is placed on the compressed spring and released from rest on a frictionless surface. This block then strikes another block of mass 2m, which is hung on a massless cord of length L as shown in the figure to the right. The masses stick together after the collision and they both swing up to a height h. Give answers for parts (a-c) in terms of k, xo, m, and/or g.

a) What is the velocity of the mass m just before the collision with the mass 2m?
b) What is the velocity of the combined masses (m+2m) just after the collision? (i.e. before they start to swing up on the cord.)
c) To what height h, do the combined masses swing up on the cord?
d) Calculate ?max, the maximum angle from the vertical position that is reached by the masses, if
you are given the following values: k = 1500. N/m2, xo = 0.100 m, m = 2.00 kg, L = 1.00 m and
g = 9.80 m/s2. Give your answer in degrees.

Explanation / Answer

a) Use the law of conservation of energy. Initial energy: -(1/2)k(xo)2, final energy: (1/2)(m)(v2), work due to nonconservative forces =0J so Eo=Ef, -(1/2)k(-(xo)2)=(1/2)(m)(v2) => (1/2)(k)(xo)2=(1/2)(m)(v2) =>

k(xo)2=m(v2) => k(xo)2/(m)=v2 =>[((k)(xo)2)/(m)]=v

(b) Use the law of conservation of momentum. Initial momentum: mv, final momentum: (m+2m)vf1 ,Impulse=0Ns so Po=Pf, mv=(m+2m)vf1 => ((mv)/(m+2m))=vf1

(c) Again use the law of conservation of energy: Initial energy: (1/2)(m+2m)(vf1)2 final energy: mgh, work due to nonconservative forces =0J so Eo=Ef, (1/2)(m+2m)(vf1)2=mgh => (1/2)(3m)(vf1)2=mgh => (1/2)(3)(vf1)2=gh =>(3/2)(vf1)2=gh => h=[((3/2)(vf1)2)/(g)]

(d) k=1500(N/m), xo=0.100m, m=2.00kg, L=1.00m, g=9.80(m/s2), and max=?

using the equation from part (a): [((1500)(0.100)2)/(2.00)]=v=2.7386(m/s)

using the equation form part (b): ((2.00)(2.7386))/(2.00+2(2.00))=vf1=.91287(m/s)

using the equation form part (c) [((3/2)(.91287)2)/(9.80)]=h=.12755m

Now set up a triangle relation: first use L as the hypotenuse, second use L-h as the adjacent, third use the inverse cosine function to find theta: =cos-1(adjacent/hypotenuse) => cos-1((L-h)/(L)) :plug in the nubmers:

=cos-1((1-.12755)/(1)) => cos-1(1-.12755) => cos-1(.87245)= 29.255o using three significant figures =29.3o

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