A steel ball of mass m 1 = 0.9 kg and a cord of length of L = 2.2 m of negligibl

Question

A steel ball of mass m1 = 0.9 kg and a cord of length of L = 2.2 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 = 0.9 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.

(a) What is the velocity of the block just after impact?

(b) How far does the block slide before coming to rest (assuming that the shelf is long enough)?

Explanation / Answer

Applying conservation of energy for the pendulum, as V^2 = 2gh

so Vf = sqrt(2 * 9.8 * 2.2)

Vf = 6.56 m/s

now from By Conservation of linear momentum,

since the masses are identical, pendulum will stop.

a) velocity of block just after impact = 6.56 m/s

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b) applying work energy theorem,

frictional force = 0.1*0.9*9.8 = 0.882 N

-0.882 * x = 0- 0.5*0.9 *6.56

Solving, x = 3.34 m

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