A block of mass 56 kg is attached to a cord of length 558 cm

Solution

The potential energy of the pendulam block is converted to kinetic energy when reaches bottom

since the pendulam block is lifter to a the radial length = 55.8 cm

the potential enrgy energy is = mgh

this potential energy is converted to kinetic enrgy ; when the block passes bottom it pocessess only kinetic enregy = mv^2/2

so mgh = mv^2/ 2

              v = sqrt (2*g*h)

              = sqrt (2*9.8 m/sec2 * 0.558m) : g acceleration due to gravity

              = 3.307 m/sec is the velocity of pendulam block when reaches bottom

in elastic collisions momeutm and velocity are conserved

i.e moemtum beofre collision m1u1 + m2u2 = mometum after collision m1v1+m2v2

m1 mass of pendulam block and u1 and v1 intial and final velocity ; m2 mass of sitting block u2 and v2 are intial and final velocitys

simillarly the kineti enrgy before collision m1u1^2/2 + m2u2^2/2 = m1v1^2 / 2 + m2v2^2 / 2

from the two equations we have

velocity of pendulam block v1 = u1(m1-m2) + 2m2u2 / (m1+m2)

                                              = 3.307 (-1.97) + 0 / 3.09 kg

                                                = -2.10834 i ; negative sign indicate negative direction of m1 after collision

velocity of floor block v2 = u2(m1-m2) + 2m1u1 / (m1+m2)

                                                  = 0 + 2 * 0.56 * 3.307 / 0.56+2.53 kg

                                                  = 1.198 in postive direction