A steel ball of mass m 1 = 1.2 kg and a cord of length of L = 2.1 m of negligibl

Question

A steel ball of mass m1 = 1.2 kg and a cord of length of L = 2.1 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m2 = 1.2 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.10. What is the velocity of the block just after impact?

Explanation / Answer

m1 = 1.2 kg L = 2.1 m

when the ball is relased from tis horzontal position, it will loose kientic energy of m1gL and this will be the gain in KE when it is at it lowest position

The velocity at the lowest position

v= SQRT(2gL) = SQRT(2*9.8*2.1) = 6.42 m/s

When it strikes m2=2.1 kg, m1 will come to rest and m2 will have a velocity 6.42 m/s

just after collison velocity of m2 is 6.42 m/s

As it starts moving kinetic friction will come into play.

Normal force N = m2g

Force of friction f = 0.1*N

Due tothis force m2 will have a retardation

retardation a = f/m2 = 0.1*g = 0.98

Velocity of m2 at any time t   Vt = 6.42 - 0.98*t

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