A cubical block of mass m is released from rest at a height h on a frictionless surface of a movable wedge of mass M, which in turn is placed on a horizontal frictionless surface.
Find the velocity of the wedge when the smaller block reaches the bottom.
Answers
Horizontally, momentum is conserved, so
0 = M*u + m*v*cosα
where u, v are the velocity of the wedge, block respectively
v = -M*u / m*cosα
and since the surfaces are frictionless, the block's initial PE is converted into KE in the block and the wedge:
m*g*h = ½(M*u² + m*v²)
Substitute for v:
m*g*h = ½(M*u² + m*(M*u / m*cosα)²) = ½(M*u² + (M²/m)u² / cos²α)
multiply through by 2/m
2*g*h = u²(M/m)(1 + (M/m)/cos²α)
u = SQRT[2*g*h*(m/M)/(1 + (M/m)/cos²α)]
which may be expressed other ways.
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Answer:
A cubical block of mass m is released from rest at a height h on a frictionless surface of a movable wedge of mass M, which in turn is placed on a horizontal frictionless surface. Find the velocity of the wedge when the smaller block reaches the bottom.
Explanation:
Horizontally, momentum is conserved, so
0 = M*u + m*v*cosα
where u, v are the velocity of the wedge, block respectively
v = -M*u / m*cosα
and since the surfaces are frictionless, the block's initial PE is converted into KE in the block and the wedge:
m*g*h = ½(M*u² + m*v²)
Substitute for v:
m*g*h = ½(M*u² + m*(M*u / m*cosα)²) = ½(M*u² + (M²/m)u² / cos²α)
multiply through by 2/m
2*g*h = u²(M/m)(1 + (M/m)/cos²α)
u = SQRT[2*g*h*(m/M)/(1 + (M/m)/cos²α)]
which may be expressed other ways.
YOU ARE WELCOME!