AP Physics Question: Linear Momentum?

Question

A 2-kg block and 8-kg block are both attached to an ideal spring (k = 200 N/m) and both are initially at rest on a horizontal frictionless surface.
A 100-gram ball of clay is thrown at the 2-kg block, moving horizontally with speed v when it hits and sticks to the block. 8-kg block is held still by a removable stop. Spring compresses a maximum distance of 0.4 m.

a. Energy stored in the spring at max. compression.
b. Speed of the clay ball and 2-kg block immediately after the clay sticks to the block but before the spring compresses significantly.
c. Initial speed v of clay

An identical ball of clay is thrown at another identical 2-kg block, but this time the stop is removed so that the 8-kg block is free to move.

d. Max. compressionn of the spring will be greater than, equal to, or less than 0.4 meter?
e. State principles used to calculate v of 8-kg block at the instant that the spring regains its original length. Write equations and show numerical substitutions.

Answer 1

a) The energy stored at max compression is
.5*k*x^2
=0.5*200*0.4^2
16 J

b) Based on the energy stored in the spring,
.5*2.1*speed^2=16
speed=sqrt(32/2.1)
speed=3.9 m/s

c) Using conservation of momentum
0.1*v=2.1*3.9
v=81.9 m/s


d) I assume that the v of the clay is 81.9
and that there is no resistance

The energy that gets transferred to the
system by the collision of the spring and
2-kg block is still 16 J, and the initial speed
of the combined 2.1-kg block and clay is still
3.9 m/s

d) The compression will be less since some energy will be transferred into kinetic energy of the 8-kg block

when there is no energy in the spring, all
of the energy in the system will be KE of the
two blocks. Also, the center of mass of the system is always moving according to the conservation of energy in the system
2.1*3.9=2.1*v1+8*v2
when the spring is elongated again
.5*2.1*v1^2+.5*8*v2^2=16

since there are two equations for v1 and v2

2.1*(3.9-v1)/8=v2
2.1*v1^2+4.41*(3.9-v1)^2/8=32
2.65*v1^2-4.3*v1-23.6=0


There are two solutions
v1=3.9 m/s
v2=0

and
v1=-2.3
v2=1.6

There are two solutions because the masses will oscillate as they move. The first elongation of the spring is the second solution. Note that the first solution is a repeat of the instant after collision of the 2-kg mass and the clay. Without friction or any outside force in the horizontal, the system will repeat the motion infinitely.


j

Answer 2

the respond is a results of viewing/understand-how a "loose-physique" diagram of the two kg block in this social gathering. for the clarification that the two kg block is to the main suitable of the three kg block, it fairly is rigidity on the three kg block could be to the left. The rigidity of the two kg block against the three kg block would be computed as as we talk as the two-block gadget's internet acceleration is computed. the internet rigidity that acts on the two block gadget = 20 N, and the completed mass of the two block gadget is 5 kg, so the internet acceleration of the two block gadget is: Anet = Fnet/5 = 20/5 = 4 m/s² for the clarification that the two blocks improve up on the comparable time, the internet acceleration of the two kg block is 4 m/s², so the rigidity that IT exerts (against 3 kg block) is F = ma = (2)(4) = 8 N.