In the figure, block 2 of mass 1.90 kg oscillates on the end...

In the figure, block 2 of mass 1.90 kg oscillates on the end of a spring in SHM with a period of 12.00 ms. The position of the block is given by $x = (0.900 \text{ cm}) \cos(\omega t + \pi/2)$. Block 1 of mass 3.80 kg slides toward block 2 with a velocity of magnitude 6.90 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time $t = 3.00 \text{ ms}$. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Transcript

00:01 You note that the spring constant k is going to be equal to 4 pi squared times m sub 1 divided by t squared.

00:11 This is going to be equal to 4 pi squared, multiplied by the block of mass 1, 4 .00 kilograms, and then divided by the period squared.

00:22 We know that the period is 20 milliseconds, so 20 times 10 to the negative 3rd, seconds, quantity squared, this is giving us 1 .97 times 10 to the fifth newtons per meter.

00:38 We can essentially it's we first need to determine when where in its simple harmonic motion block where is block two when the impact occurs in its simple harmonic motion.

00:57 We can say that the we know the frequency, the angular frequency rather, is equaling 2 pi divided by the period t.

01:06 And so we can say that the given value of t when the collision takes place is one -fourth of the period.

01:16 So we can then say that omega -t is equaling pi over 2.

01:23 And the location then of block 2, we can say location of block 2, would be x.

01:34 This is going to be equal to the maximum displacement times cosine of omega -t plus the face constant phi.

01:43 We know that phi here in this case is equaling pi over 2.

01:54 This gives us that x is equaling x -sub -m, the maximum displacement times cosine of pi over 2 plus pi over 2.

02:07 And this is equal in pi over two plus pi of two is equal pi.

02:11 Cosine of pi is negative 1.

02:13 So negative 1 times the maximum displacement is going to be equal to negative x of m, negative maximum displacement.

02:20 So essentially, this means that block 2 is at its turning point, given that it is at the negative maximum displacement when the impact occurs.

02:40 So this also means that the string is stretched to an amount of one centimeter at this moment.

02:47 So we can say that at this moment, string is stretched an amount equaling 0 .01 meter or one centimeter.

03:16 So we can then say that to calculate its after collision speed, we're going to use momentum conservation.

03:27 So at this point we can say that v, the final speed after the collision is going to be equal to 4 .0 kilograms.

03:37 This would be multiplied by 6 .0 meters per second...

Question

Please give Ace some feedback