1) A block of mass m = 3.6 kg, moving on a frictionless surface with a speed Vi=

Question

1) A block of mass m = 3.6 kg, moving on a frictionless surface with a speed Vi=9.3m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 3.6 kg block recoils with a speed of Vf=2.7m/s. What is the speed of the block of mass M after the collision?


2) A 50 kg uniform ladder, 5.0 m long, is placed against a smooth wall at a height of h = 3.7 m. The base of the ladder rests on a rough horizontal surface whose coefficient of static friction is 0.40. An 80 kg block is suspended from the top rung of the ladder, just at the wall. What is the magnitude of the force exerted on the base of the ladder, due to contact with the rough horizontal surface?

Explanation / Answer

a) by applying momentum of conservation

mvi = mvf + Mvf'

3.6(9.3) = 3.6(-2.7) + Mvf'

Mvf'= 43.2

now coefficient of resitution = 1

vi - vi' = vf' - vf

9.3 - 0 = vf' - (-2.7)

vf' = 6.6m/s --ans

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