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Two small spheres of putty, A and B, of mass M1 = 0.25 kg and M2 = 0.30 kg, resp

ID: 2204836 • Letter: T

Question

Two small spheres of putty, A and B, of mass M1 = 0.25 kg and M2 = 0.30 kg,
respectively, hang from the ceiling on strings of equal length. Sphere A is drawn aside
and raised to a height hi = 0.15 m as shown and then released. Sphere A collides with
sphere B. They stick together and swing to a maximum height hf.

(a) [4 points] If the zero reference of the gravitational
potential energy is chosen to be at the bottom of the
swing where B is located before collision, what is the
gravitational potential energy of A at the time of release?
__________

(b) [4 points] What is the velocity of A at the instant it collides with B?
__________


(c) [4 points] What is the common velocity of A and B immediately after collision
__________


(d) [4 points] What fraction (or percentage) of the initial mechanical energy remains in
the two-mass system after the collision?
__________


(e) [4 points] What is the maximum height hf reached by A+B after collision?
__________




Explanation / Answer

a). gravitational energy at point A= m1ghi = 0.25*9.81*0.15 = 0.3679 J

b).

velocity, v1 of A at the instant it collides with B:

from the energy equation,

m1ghi = (1/2)*m1*v12

v1 = (2ghi) = (2*9.81*0.15) = 1.716 m/s

c).

common velocity,v2 of A and B immediately after collision:

consider elastic collision. from the conservation of the linear momentum,

mv= constant

m1v1 = (m1+m2)*v2

therefore,

0.25*0.1716 = (0.25+0.3)*v2

v2 = 0.8 m/s

d).

fraction (or percentage) of the initial mechanical energy remains in
the two-mass system after the collision = initial mechanical energy/energy after collision

= [(1/2)*(m1+m2)*v22]*100/[m1ghi]

= [(1/2)*(0.25+0.3)*0.82]*100/(0.25*9.81*0.15) = 47.84 %

e).

from the energy conservation,

(1/2)*(m1+m2)*v22=(m1+m2)*g*hf

(1/2)*(0.25+0.3)*0.82=(0.25+0.3)*9.81*hf

hf= 0.0326 m

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