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## Homework Statement

In the figure, block 2 of mass 2.40 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (1.80 cm) cos(ωt + π/2). Block 1 of mass 4.80 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 = 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

## Homework Equations

E=KE+PE=0.5k*a^2 where k is spring constant and a is amplitude

KE=0.5m*v^2

T=2π*sqrt(m/k)

## The Attempt at a Solution

First rearrange the expression for T:

k=(4*m*π^2)/T^2

We also have initial mechanical energy:

Ei=0.5k*a^2, where a can be read off from the given equation x = (1.80 cm) cos(ωt + π/2), that is 0.018m, and we have managed an expression for k.

Thus we get Ei=45.42286025 J

After the collision, block 1 transfers all its energy to block 2, thus the total mechanical energy increases by the the KE of block 1:

KE1=0.5*M*v^2=0.5*4.8*6.9^2=114.264 J

Now the new mechanical energy is

Ef=Ei+KE1=159.6758603 J

Finally we use the equation E=0.5k*a^2 to solve for the new amplitude and get

a=0.0477 m

This is not the correct answer and I did not use all the information given(t=6.5ms and ø=π/2).

Please point out what I have missed thanks.