Our system is composed of a block with mass m which is moving with the initial s
ID: 1283603 • Letter: O
Question
Our system is composed of a block with mass m which is moving with the initial speed v0
to the right on the
at part of the ramp with mass M. The ramp is allowed to move on a
at frictionless surface. We will also assume that friction between the block and the ramp is
negligible. The question is:
What is the velocity of the ramp at the moment the block leaves the ramp?
Express your answer in terms of v0 and H.
Lets denote the velocity of the ramp and the block, both respect to the ground frame, as
~V= V^i, and ~v = vx^i+ vy^j
.
1. What is the initial kinetic energy?
2. What is the initial momentum vector of the system? (use ~A= Ax^i+ Ay^jnotation.)
3. Is the mechanical energy of the system a conserved quantity during the motion? (if yes.
please elaborate)
4. What are the external forces on the system (block + ramp)? What is the x-component
of the net force? (it is zero. Please explain).
5. Is momentum in y-direction conserved? (No. Please elaborate.)
6. Use conservation of the mechanical energy and conservation of momentum in x-direction
to nd V and vx;y in terms of m, M, v0, and g.
Explanation / Answer
1)KE initial=0.5mv0^2
2)Momentum=mvo i
3)Mechanical energy is conserved since no losses due to friction and gravity is a conservative force.
4)External forces are gravity and normal reaction from the surface to the ramp. Both are in y direction. No force acts in the x-direction since there is no friction.
5) momentum in y-direction is not conserved since it is 0 initially but not zero finally since the block of mass m has some final y-direction velocity. This is due to the work done on the system by the normal reaction.
6)Conserving momentum in x-direction.
mv0=(M+m)vx
vx=mv0/(M+m)
This is because the horizontal velocities of both the ramp and the block are equal at the point of separation.
Conserving mechanical energy,
0.5mv0^2=mgH+0.5*(m+M)vx^2+0.5mvy^2
0.5mv0^2-0.5*(m+M)(mv0/(m+M))^2-mgH=0.5mvy^2
v0^2-mv0^2/(m+M)-2gH=vy^2
vy^2=v0^2(1-m/(M+m))-gH=M/(m+M)-gH
vy=sqrt(M/(m+M)-gH)
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