Can someone work these two problems out for me? 9) A pendulum bob that is attach

Question

Can someone work these two problems out for me?

9) A pendulum bob that is attached to a 2.00-m cord is pulled sideways until the cord makes an angle of 36.9 with the vertical. If the bob is now released from rest, find its speed as it passes point. through its lowest (10) A block of mass 250.0 g moves across a frictionless surface with a speed of 3.00 m/s. It makes a completely inelastic collision with a second block whose mass is 150.0 g, and which is at rest. What is the final velocity of the joined blocks?

Explanation / Answer

initial mechanical energy Ei = m*g*L*(1-costheta)

final mechanical energy Ef = (1/2)*m*v^2

from energy conservation

Ef = EI

(1/2)*m*v^2 = m*g*L*(1-costheta)

v = sqrt(2*g*L*(1-costheta))

v = sqrt(2*9.8*2*(1-cos36.9))

v = 2.8 m/s

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10)

from momentum conservation

momentum before collision = momentum after collision

m1*v1i + m2*v2i = m1*v1f + m2*v2f

since collision is Inelastic

v1f = v2f = v

(250*3) + (150*0) = (250 + 150)*v

v = 1.875 m/s <<<<<<==========ANSWER

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