Question

Chapter 15, Problem 034 GO In the figure, block 2 of mass 2.30 kg Oscillates on the end of a soring in SHM with a period of 20.00 ms. The position of the block is given byx (0.500 cm) cos( 2), Block 1 of mass 4.60 kg slides toward block 2 with a velocity of magnitude 6.60 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at timet5.00 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM alter the collision? Number Units the tolerance is +/-5%

Answer 1

T= 20.00ms, o = 25 t= 5.00ms and 3 = (0.500 cm) cos (cot + 1/2) therefore at t = 5:00ms x = -5.500cm that is at extreme position Hence at t=5.00 ms when collision occurred block 2 have no velocity. Conservation of momentum mu, + MzV2 = (m, + m₂) mi = 4.60 kg, m2 = 2:30 Kg,Vi = 6.60 m/s, Vz=0 (at t = 5:00ms it is at extreme position, hence V2 =o) Final common speed, v= m. Vi (m, tm2) =) v = 4.40 mis Let New amplitude be a, then by energy conservation 1 (mitm) V2 + xač = į kai where K= mw² = my 47² = 2.27X105 N/m (spring constant) 7² and ao = 0.500 cm = 5.00x10-3 m (original Ja, = 2.49cm) amplitude)