The rectangular block (m2 = 4 kg) is free to roll along the horizontal guide. A

Question

The rectangular block (m2 = 4 kg) is free to roll along the horizontal guide. A ball (m1 = 2 kg) is attached to the block with a rigid massless rod at point O through a frictionless pivot. The system is released from rest when theta = 45. How far wil the block have moved when theta = 0, and how fast will it be moving. L = 3m.  ANSWER: 0.7071 m and 1.695 m/s

The rectangular block (m2 = 4 kg) is free to roll along the horizontal guide. A ball (m1 = 2 kg) is attached to the block with a rigid massless rod at point O through a frictionless pivot. The system is released from rest when theta = 45. How far wil the block have moved when theta = 0, and how fast will it be moving. L = 3m. ANSWER: 0.7071 m and 1.695 m/s

Explanation / Answer

a)both move in opp directions ......the m2 moves along right

conservation of momemtum we have

m1v1=m2v2

v1=2v2

as the ball travels from 45 to 0 change in height = L(1-cos45)

change in potential energy= kinetic energy

m1*g*L*(1-cos45)= 0.5*m1*v1^2+0.5*m2*v2^2

v1=2v2

4v2^2+2v2^2 = 2*9.8*3*(1-0.707)

v2^2 = 2.873

V2=1.695 m/s

let x be the distance covered then....total length along x direction covered by m1= Lsin45-x............. by conservation of energy we have

m1*a*(Lsin45-x) = m2*a*x

x(m1+m2)=m1*L*sin45

x= 2*3*sin45/(2+4)=sin45 =0.707 m

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