1. The problem statement, all variables and given/known data 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? 2. Relevant 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) 3. 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.