A block of mass .56 kg is attached to a cord of length 55.8 cm. The block is rai

Question

A block of mass .56 kg is attached to a cord of length 55.8 cm. The block is raised up "pendulum style" until the court is horizontal and then it is released. At the very bottom of the blocks pendulum pass it hits another block of mass 2.53 kg sitting at rest on frictionless floor. assuming that the collision is elastic and that the initial velocity of the pendulum block is in the positive direction find:
A.) The velocity of the pendulum block just after the collision.
B.) The velocity of the floor block just after the collision. A block of mass .56 kg is attached to a cord of length 55.8 cm. The block is raised up "pendulum style" until the court is horizontal and then it is released. At the very bottom of the blocks pendulum pass it hits another block of mass 2.53 kg sitting at rest on frictionless floor. assuming that the collision is elastic and that the initial velocity of the pendulum block is in the positive direction find:
A.) The velocity of the pendulum block just after the collision.
B.) The velocity of the floor block just after the collision.
A.) The velocity of the pendulum block just after the collision.
B.) The velocity of the floor block just after the collision.

Explanation / Answer

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

                              

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