Question

A dog sled is sliding down a hill. The sled starts from a height of 25 m and has a weight of 66 kg. It collides with a person at the bottom of the hill. The person weighs 72 kg. The collision is completely inelastic. Friction can be ignored due to icy conditions. Determine the speed of the sled and the person after the collision Show all steps for full marks.

Answer 1

Mass of dog sledge,ms= 66 Kg

Mass ofthe person,mp=72 Kg

Kinetic energy of sledge at height 25m=0 (because initally sledge ia at rest)

Let Kb be the kenitic energy at bottom.

And potential energy at bottom=0.

therefore,

Potential energy of sledge at the at height 25m =mgh = 66*9.81*25=16186.5 J

since friction is absent using law of conservation of energy for energy at height 25 m and at bottom we write

0+16186.5 = Kb+0

or Kb=16186.5.

However

Thus the speed of sledge before collision (Vsi )is 1461.71 m/s.

Applying Law of conservation of momentum to this perfectly inellstic  collision we must have:

with vsi= 1461.71 m/s vpi=0 (person was initially at rest ) we may write

Hence speed of the sled and the person after the collision is 699.07 m/s