Question

A bull with a weight of 500-kg runs at a speed of 15 m/s and hits at a standing idle 70-kg weighing man. After impact, the idle man has thrown at speed of 5 m/s and the bull has slowed down its speed to 10 m/s. What would be the sum of the kinetic energy (in Joules) of the bull and man after impact by the bull, assuming that the bull and the man as a system and the collision is not an elastic one?

Answer 1

After collision :

K . E. = m v^ 2 / r

K . E . = 500 x 15 ^2 / 2

K . E . = 56250 Joule

Before collision

K . E. = m v^ 2 / r

K . E . = 70 x 5 ^ 2 / 2

K . E . = 875 Joule

K . E . = 500 x 10 ^ 2/ 2

K . E . = 25000 Joule

Total K . E . = 25000 + 875 = 25875 Joules