Question

Problem 4. A child sleds down a hill and collides at 5.6 m/s into a stationary sled that is identical to his. The child is launched forward at the same speed, leaving behind the two sleds that lock together and slide forward more slowly. What is the speed of the two sleds after this collision

Answer 1

Answer:

Explanation:

mass of both the sleds = m

initial velocity of first child = u

initial velocity of second sled = 0

Let v be the speed of both the locked sleds after collision.

Use the conservation of momentum

Mass of first sled x initial velocity of first sled + Mass of second sled x initial velocity of second sled  = 2 x mass of one sled x final velocity of locked sleds

m x u + mx 0 = 2m x V

mu = 2 mV

V = u/2 = 5.6 /2 = 2.8 m/ s

Thus, the speed of the sleds after collision is 2.8 m/s .