Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. you (inertia 80 kg ) then move across horizontal snow toward a 400-kg boulder but jump off the sled (inertia 8.0 kg ) the instant before it hits the boulder. the boulder is sitting on very slick ice and moves freely when the sled hits it. the sled bounces back, moving at 6.0 m/s. at what speed does the boulder move after the sled hits it?

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128  kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y