Horizontal motion, collisions, springs?

Question

A block of mass "m" is moving on a horizontal frictionless surface with a speed of "v sub 0" (initial velocity) as it approaches a block of mass "2m" which is at rest and has an ideal spring attached to one side, (the side where the impact will occur). When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.

A.) Briefly explain why the two blocks have the same speed when the spring is completely compressed.

B) Determine the speed "v sub f" (the final velocity) of the two blocks while the spring is completely compressed.

C) Determine the kinetic energy of the two blocks as they move together with the same speed.

D) Write the equations that could be used to solve for the speed of each block after they have separated. It is not necessary to solve the equations for the two speeds.

Answer 1

Muten's answer is incorrect since it is an inelastic collision due to the spring compression.

This problem is useful in understanding elastic versus inelastic collisions. Note that the spring models the loss of energy in an inelastic collision, which gets converted to potential energy in the spring in this case. Inelastic collisions are almost always where the colliding objects stick together.

A. When the spring is completely compressed, the two blocks are essentially connected together. Looking at a free-body diagram they must have the same speed, and in this case the same velocity (velocity is a vector).

B. Using conservation of momentum:

mvf+2mvf=mvo
or vf=vo/3

C. The kinetic energy is:
.5*(m+2m)*vf^2
=.5*3m*vf^2

Note that this does not equal the kinetic energy prior to collision.
.5*m*vo^2
using the vf from B:
(.5*3m*vo^2)/9
or
(.5*m*vo^2)/3

2/3 of the kinetic energy was converted into other forms of energy after collision.

For D, consider free body diagrams of the blocks the instant the spring is fully compressed.
Since there is no friction, the only horizontal force on the blocks is the spring between them.
For mass 2m, the force is k*L
and for mass 1m, the force is -k*L
where k is the spring constant and L is the length of compression.

F=m*a
so a=F/m
As the blocks separate, the force from the spring will decrease at a rate of F-k*dx where dx is the incremental uncoiling
since a=F/m
for block m
a=-k/m*(L+dx) from x=L to zero
and for block 2m
a=k/2m*(L-dx) from x=L to zero

Using calculus this could be solved if you know the k and L

If you assume conservation of energy, which implies that the spring was long enough to compress without loss, then the energy in the spring was equal to the loss of kinetic energy in the collision. Under this assumption, you can model that the final kinetic energy will be equal to the starting kinetic energy as if it was an elastic collision. Or
.5mvo^2=.5mv1^2+.5*2m*v2^2
or
vo^2=v1^2+2v2^2
also, momentum will be conserved:
mvo=mv1+2mv2
or
vo=v1+2v2

This will produce a quadratic equation to find the vectors v1 and v2 that satisfy both equations.

j

Answer 2

When the first mass touches the spring, the spring starts getting compressed. The spring force acts to retard the first mass and accelerate the second mass. The speed of the first mass starts decreasing whereas that of the second mass increases. As long as the speed of m1 is greater than that of m2, the spring willkeep on getting compressed more and more. The max compression occurs when both the masses are moving with the same speed.

This speed vf can be found out using conservation of momentum of the system comprising of the two masses and the spring.
The initial momentum = m1*v0
The momentum when the spring is at max compression = m1*vf + m2*vf.
This gives vf = v0*m1/(m1+m2).

The KE's of the two masses at this time are of course given by 1/2 m1*vf^2 and 1/2 m2*vf^2.

Let the speeds of the two masses be v1 and v2 when they finally separate. Since the whole process has been conservative, the initial KE of the system, namely 1/2 m1*vo^2 must be equal to the final energy of the system. Since the spring is no more under compression, it does not store any energy and the final energy of the system = 1/2 m1*v1^2 + 1/2 m2*v2^2.This gives one equation. The other of course is given by the conservation of momentum, that is
m1*vo = m1*v1 + m2*v2.
Using these two equations you can find out the two unknowns v1 and v2.
odu has facilely concluded that my answer is wrong since he claims it is an inelastic collision. Please note that I have described the system as including the spring, in which case it does become an elastic collision. As a matter of fact, odu has also come to the same conclusion as mine.

Answer 3

a. PE = mgh Ff = Friction Force = 300N Wf = Work done by friction Wf= Ff *h KE = PE - Wf 1/2 m v^2 = mgh - 300N * h 1/2 m v^2 = h (mg - 300 N) v = √ (2h (mg - 300 N)/m v = √( 2 * 4.5m (56 * 9.8 - 300 N)/56) v = 6.32m b. The easy answer is simply have KE at bottom equal spring compression, but there is a better answer. Spring Energy = KE.bottom + gravity PE for d distance - Ff *d 1/2 k d^2 = 1/2 m v^2 + mgd - Ff d 0 = kd^2 + 2d (Ff -mg) - mv^2 0 = 2300N/m d^2 - 497.6J/m d - 2239J Solve Quadratic d = 1.10 m ( or -0.88 m which is rejected)