Question

4. Two blocks with masses 10.0 kg and 15.0 kg are placed on a horizontal frictionless surface. A light spring is placed in a horizontal position between the blocks. The blocks are pushed together, compressing the spring, and then release from rest. After blocks have lost contact with the spring, the 10.0 kg mass has velocity, v = 4.0 î m/s. (a) How much potential energy was stored in the spring before the blocks were released? (b) What is the final velocity of the center of mass of the two-block system?

Answer 1

mus @ given, mass of sof block, mass of sof block, m, = 10.0 mg of and block, 15. okey velocity of block 1 of the last of contact velocity of block 2, offer last Comentum theoen Total monendum before release & tokal momentan offer release V= 4 m/ que alory 1 u-asip Lef of contact : Then as per the censen nation of entorno mv, + m₂ V₂ Toted monendum Refore release to zero as all syskon week in weer in rest, oy velocity of blocks, after release V25 mui 2 9.667 mio - (long (4m1B) (25mg) [" the potential energy rates dored in the exprong before the block were released given by PE= m 2 + 2 m, g2 1 10kg) (4m² + 12 (15kg) (-2.667 m)? same as the kinetic enengy of the system as per conservation of enengy 7 PE - 133. 347 J ③ The the velocity of the cele oras of the two back typ fem å given by 4 min + m₂ uz Vem Crong)(4) 4(151g)(2007) -3.333X16 ( lovg) 15kg) Vm= -3, 33x10 4 mip m,+ m₂

Comment

The negative sign indicates the velocities indicates that, the velocity is along negative x axis.

So the velocity of centre of mass of the two block system in vector format is given by

Vcm = -3.33*10^-4i m/s