In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in
ID: 1494626 • Letter: I
Question
In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos(t + /2). Block 1 of mass 2.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?
Answer: _________ m
Explanation / Answer
The completely inelastic collision means that linear momentum will be conserved but kinetic energy will be dissipated as heat.
We're going to need the spring constant "k", so let's get that right away.
T = 2 pi sqrt(m/k) = 0.020 s = 2pi*sqrt(2.0/k)
(0.020 s)^2 = 4 pi^2 (2/k )
k = 8pi^2/(0.020)^2 = 197392.1 N/m
Since you don't show the figure, I'm not sure of my interpretation, but it sounds like block 1 maybe has a hole in it and is sliding right down the spring. (?) In the +x- direction?
OK, at t = 5.0 ms, the Block 2 has gone through 1/4 of a cycle, but at t=0 its displacement was 0
(by the equation), and after 1/4 cycle, its displacement is -1.0 cm and its speed is zero.
The near-instantaneous effect of the collision with Block 1 (which is moving TOWARDS the equilibrium point of the stretched spring) is to change the speed of the combined mass.
The (6.0 m/s)(2.0 kg) momentum of the Block 1 must now be "spread" over Blocks 1 and 2,
whose combined mass is 6.0 kg, so the near-instantaneous effect is that the speed of the
combined mass becomes (1/3)(6.0 m/s) = 2.0 m/s.
We now need to figure out the new SHM period, which should be 2pi*sqrt(6.0 kg / k ) = the old period times sqrt(3) = 3.644 ms.
If the "x" direction is horizontal, there is no reason to suppose that the equilibrium position
has changed. The new motion will be x = A cos(2 pi t/34.64 ms + phi),
and the maximum speed will be 2.0 m/s PLUS some further speed imparted by the
action of the spring, The kinetic energy at the collision position (-1.0 cm) after the
collision is (1/2)(6.0 kg)(2.0 m/s)^2 = 12.0 J, and the potential energy of the spring at this position is
(1/2)( 197392.1 N/m)(0.010 m)^2 = 9.87 J.
Thus, the total energy of the SHM after the collision should be 21.87 J,
and this is the (1/2)kA^2 where A is the new amplitude.
(21.87 J) = (1/2) (197392.1 N/m) A^2
A = 1.49 cm