Question

A block of mass m 2.20 kg slides down an incline which is 3.60 m high. At the bottom, it strikes block mass M 7.00 kg which at rest on a horizontal surface, as 3.60 m shown (assume a smooth transition to the bottom of the incline). If the collision is elastic, and friction can of mass m 2.20 kg just before its strikes the block be ignored, determine (a) the speed of the block of mass M-7.00 kg. (b) the speeds of the two blocks after the collision.

USE these formulas { Va^2=(2*g*h)^2 }

and { Ma*Va^2+Mb*Vb=Ma*Va'+Mb*vb' }

and { 1/2Ma*Va^2+1/2Mb*Vb^2=1/2MaVa'2+1/2Mb*Vb'^2 }

Answer 1

let mass of the first body m₁=2.20kg

         height of inclination h=3.60m

mass of second body     m₂=7.00kg

the second body is at rest.so u₂=0

the speed of first body u₁²=2gh

                                      =2x9.8x3.60

                                   u₁ =8.4m/sec

the speed of the first body after collision v₁=(m₁-m_2/m₁+m₂)u₁

                                                              = (2.20-7.00/2.20+7.00)8.4

                                                               =-4.382m/sec

the speed of the second body after collision v₂=(2m₁/m₁+m₂)u₁

                                                                                     =(2x2.20/2.20+7.00)8.4

                                                                 =4.017m/sec