8. 15 points Tipler6 8.P.103 My Notes Ask Your Teacher A steel ball of mass m1 =

Question

8. 15 points Tipler6 8.P.103 My Notes Ask Your Teacher A steel ball of mass m1 = 0.9 kg and a cord of length of L = 1.9 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 owest 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 M1 (a) What is the velocity of the block just after impact? m/s (b) How far does the block slide before coming to rest (assuming that the shelf is long enough)?

Explanation / Answer

Given,

m1 = 0.9 kg ; L = 1.9 m ; m2 = 0.9 kg ; uk = 0.1

a)from conservation of energy the speed of the ball at the lowest point is:

v = sqrt (2 g L)

v = sqrt (2 x 9.81 x 1.9) = 6.11 m/s

from conservation of momentum

Pi = Pf

0 = m1 v1 + m2 v2

v2 = -m1 v1/m2

v2 = 0.9 x 6.11/0.9 = 6.11 m/s

Hence, v2 = -6.11 m/s

B)KE = Wf

where Wf is the work done by the frictional force

1/2 m v^2 = uk m d g

d = 0.5 v^2/uk g = 0.5 x 6.11^2/0.1 x 9.81 = 19.03 m

Hence, d = 19.03 m

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