a block of mass m is released from the top of a wedge of mass M. find disp. of M after m reaches the bottom.?

Question

height of wedge h. & inclination theeta

neglect friction everywhere

Answer 1

You have two masses that will be set into motion - the block sliding down the plane (wedge) and the wedge sliding due to the blocks motion.

Consider the block first, you need to figure out how long it takes to reach the bottom of the wedge (ignore that the wedge moves). The force along the plane is m·g·sinΘ for the block.

F=m·a
a=F/m
a=(m·g·sinΘ)/m
a=g·sinΘ

We will call the distance that the block slides "d". Use formula:

x'-x = v·t+1/2·a·t^2
where v = initial velocity = 0
x'-x = displacement = d

solve for t to get t=√((2·d)/(g·sinΘ))

Now let's look at the wedge of mass M, it will have a force applied to it from the Normal force of the wedge. For a vertical component of F=mg, there will be a horizontal component of F=m·g·tanΘ. This is horizontal component of the Normal force acting on the wedge, and will cause the wedge to accelerate at

a=F/M
a=(m·g·tanΘ)/M

From the previous equation, we can get the displacement of the wedge:

x'-x = v·t+1/2·a·t^2
x'-x = 1/2·(m·g·tanΘ/M)(2·d/(g·sinΘ))
x'-x = m·d·tanΘ/(M·sinΘ)

where d=h/sinΘ
x'-x=m·h·tanΘ/(M·sinΘ^2)

This is the displacement of the wedge when the block reaches the bottom.

m

Answer 2

Sorry, your question makes no sense. What does "disp. of M" mean? My guess is you mean dissipation of kinetic energy when the mass m hits bottom.

Unfortunately, when you assumed no friction everywhere, you did away with the one thing that would dissipate that energy as heat. Thus, in your frictionless case, once PE = KE; where PE = potential energy and KE = kinetic energy, that block will continue to slide until it runs into something or goes back up to h and becomes PE once more.

OK, given we know now that the mass will slide on forever, there are some things we can find out about it. First, KE = PE from the conservation of energy law. Thus, PE = mgh = 1/2 mv^2 = KE; so that v = sqrt(2gh) and we can find out how fast that little guy is moving along the floor. Note the physics, its velocity (v) does not depend on its mass in any way. Nor does it depend on that theta you gave because of the frictionless world you proposed.