Mass M1 = 2.00 kg is at the end of a rope that is 2.00 m in length. The initial

Question

Mass M1 = 2.00 kg is at the end of a rope that is 2.00 m in length. The initial angle with respect to the vertical is 60.0° and M1 is initially at rest. Mass M1 is released and strikes M2 = 5.00 kg exactly horizontally. The collision is elastic. After collision, mass M2 is moving on a frictionless surface, but runs into a rough patch 2.00 m in length, with coefficients of friction = 0.100 for both kinetic and static. At the end of the rough patch, the object is again on a frictionless surface and is stopped by a spring with spring constant equal to k = 50.0 N/m. Calculate the following: a) Velocities of Masses M1 and M2 after collision b) The amount the spring is compressed while stopping mass M2. c) What maximum height after collision does mass M1 obtain?

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Explanation / Answer

a )

the velocity is u1 = ( 2 g L ( 1 - cos ) )1/2

u1 = ( 2 X 9.8 X 2 ( 1 - cos60 ) )1/2

u1 = 4.431 m/sec

using conservation of momentum

m1 u1 + m2 X 0 = m1 v1 + m2 X v2

2 X 4.43 = 2 v1 + 5 v2 ----------- 1

v2 - v1 / u1 = 1

v2 - v1 = 4.43 m/sec ------ 2

by solving the equations

v1 = - 1.9 m/sec

v2 = 2.53 m/sec

V2 ' = ( V22 - 2 X 0.1 X 9.8 X 2 )1/2

V2 ' = 1.58 m/sec

b )

1/2 k x2 = 1/2 m2V2 '2

50 x2 = 5 X 1.582

x = 1/2 m

the spring is compressed while stopping mass M2 is x = 1/2 m

c )

again using conservation of energy

1/2 m1 V12 = m1 g h

h = v12 / 2 g

h = 1.92 / 9.8 X 2

h = 3.61 / 19.6

h = 0.18418 m

maximum height after collision does mass M1 obtain is h = 0.18418 m

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