Paragraph: A horizontal spring block system executes SHM with amplitude A =10 ~cm initial phase =0 and angular frequency . The mass of block is M=13 ~kg and there is no friction between the block and the horizontal surface. The spring constant being 2500 ~N / m . At t = t ₁ [for which t ₁= ₁=30^ ]. A mass m =12 ~kg is gently put on the block. [Assume that collision between the block and the mass is perfectly inelasti

Step-by-step solution

aligned & = k M , ^ = k M+m = 2500 25 =10 rad / sec & ^ = M M+m & x₁=A ₁ aligned Potential energy stored in the spring at (t=t₁ )= 1 2 k x₁^2= 1 2 M ^2 ^2 ₁ Let v and v ' be the velocity of system just before collision and just after collision, so using COLM v^ = M v (M+m) = M A (M+m) Total energy after collision = PE + KE = 1 2 kx ₁^2+ 1 2 ( M + m ) v ^ 2 aligned & = 1 2 M A^2 ^2 [ M+m ^2 M+m ]= 1 2 k A^2 [ M+m ^2 M+m ]=8 J & 1 2 (M+m) A^ 2 ^ 2 = 1 2 M A^2 ^2 [ M+m ^2 M+m ] aligned