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 (6-34°). This sphere rolls down and collides with another solid sphere Bof 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 After Collision a) Determine the speeds of the sphere A just before it collided with sphere B b) Determine the speeds and directions of motion of the spheres Aand 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

(b)

after collision

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

the sphere A moves with 1.355 m/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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