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Question

A small block of mass M is released from rest at the top of the curved frictionless ramp. The block slides down the ramp and is moving with a speed of 3.5v when it collides with a larger block of mass 1.5M at rest at the bottom of the incline. The larger block moves to the right at a speed 2v immediately after the collision. Express the answers in terms of the variables given and fundamental constants.

a) Determine the h of the ramp from which the small block was released.
b) determine the speed of the small block after the collision.
c) The larger block slides a distance D before coming to rest. Determine the value of the coefficient of kinetic friction between the block and the surface.

please show me steps an formulas, i am so confused.

Answer 1

a) The height of the ramp can be found using conservation of energy: The kinetic energy of the block is related to the loss in potential energy as:
.5*M*(3.5v)^2=M*g*h
or
h=.5*(3.5v)^2/g

b) The collision of the two blocks must obey conservation of momentum no matter if it is elastic or inelastic.
M*3.5v=M*s+1.5M*2v
note that velocity is a vector and I chose right as positive
s is the velocity of the small block after collision
note that M divides out
3.5v=s+3v
s=.5v

c) this can be solved using energy. The kinetic energy of the larger block starts out at
.5*1.5*M*4*v^2
and the work done by friction is
f*D
where f is the force of friction, so
f*D=.5*1.5*M*4*v^2
or
f*D=3*M*v^2
since the kinetic energy all get lost due to frictional work.
The force of friction is
f=M*g*u
where u is the coefficient of kinetic friction, so
M*g*u*D=3*M*v^2
solving for u:
u=3*v^2/(g*D)

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