4. Solid sphere A of mass m1=5.0 Kg is released from rest from a height of h = 1

Question

4. Solid sphere A of mass m1=5.0 Kg is released from rest from a height of h = 1.5 m on top of a slope (?=34o). This sphere rolls down and collides with another solid sphere B of mass m2=3.0 Kg resting on the horizontal surface. After an elastic collision, they are rolling on the horizontal surface as shown. Soon after, sphere B hits a massless spring on the right.

b)Determine the speeds and directions of motion of the spheres A and B right after they collide.

c) If the spring constant is k = 2800 N/m, what is the maximum compression of the spring due to collision with block B.

Explanation / Answer

A) using law of conservation of energy

energy at the top of the incline = energy at the bottom of the incline

m*g*h = 0.5*m*v^2

v =sqrt(2*g*h)

v = sqrt(2*9.81*1.5) =5.42 m/s is the answer for part a)

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B) after collision

vA = (mA-mB)*uA/(mA+mB) = (5-3)*5.42/(5+3) = 1.355 m/s

the sphere A moves with 1.355m/s in the same direction

vB = 2*mA*uA/(mA+mB) = 2*5*5.42/(5+3) = 6.775 m/s

C) using law of conservation of energy

0.5*mB*vB^2 = 0.5*K*x^2

0.5*3*6.775*6.775 = 0.5*2800*x^2

x = 0.22176 m = 22.176 cm

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