Question

An object of mass m1=6.1 kg moving at 5.1 m/s strikes a stationary second object of unknown mass. After an elastic collision, the first object is observed moving at 3.06 m/s at an angle of -43° with respect to the original line of motion. What is the energy of the second object?

Answer 1

Given m1 = 6.1 kg

Initial velocity of the of an object having mass m1 is v = 5.1 m/s

Let the mass of the second object be M

Initial velocity of the second object V = 0

So, total energy before collison K =(1/2)m1 v ² + (1/2) MV ²

                                                 = (1/2) m1 v ²     Since V = 0

                                                 = 0.5 x 6.1 x 5.1 ²

                                                = 79.3305 J

Energy of the first object after collision k = (1/2) mv ' 2

Where v ' = Speed of the first object after collision

              = 3.06 m/s

Substitute values you get k = 0.5 x 6.1 x 3.06 ²

                                         = 28.55898 J

From law of conservation of energy , K = k + k '

Where k ' = Energy of the second object

So, k ' = K - k

          = 50.77152 J