Question

1. Ryan is readying for a face-off at the begninning of a hockey game. What is the total Force that ryan exerts on the ice? What is the total force of the ice on Ryan? Assume that Ryans total mass is 100 kg. Also take up ot be the +y direction, and towards the opponents goal to be +x direction.

2. Ryan is skating down the ice at 5 m/s. What is Ryan's momentum? Kinetic Energy?

3. Preparing to take a shot at the net, Ryan comes to a stop on the ice. If it takes ryan .5 seconds to come to stop, what is the net force that the ice exerts on Ryan? Use the net force to find the distance it takes ryan to come to a stop. What is the net force that Ryan exerts on the ice while stopping: give a name to this force.

4. Ryan takes a shot, sending the .2 kg puck towards the goal at 30 m/s. Assuming the collisions with the net are perfectly inelastic, and the collisions with the goalies stick are perfectly elastic, and that the goalie and net both remain at rest after the collision calculate:

a. The change in kinetic energy and momentum if the goalie blocks the puck with his stick

b. The change in KE and momentum if the puck goes into the net and stops.

Answer 1

1) weight of ryan = mg = 100* 9.8 =980 N

force exerted by ryan on ice = -980 N

and due to reaction force exerted by ice on ryan = 980 N

2) momentum = mv = 100* 5 = 500 kg m/s

kinetic energy = mv^2/2 = 100* 5* *5 /2 = 1250 J

3) acceleration = v-u/t = 5/0.5 = 10 m/s^s

force = ma = 100*10 =1000 N

this force is called force of friction

work = energy = 1250

F s =1250

so s =1250/1000 = 1.25 m

4) initial velocity of puck = 30m/s

initial momentum =0.2* 30 = 6kg m/s

if goalie stops the pluck

velocity of puck after collison = - velocity of puck before collsion = -30m due to elastic collision

final momentum = 0.2* -30 = -6kg m/s

change in momentum = 6 -(-6) = 12kg m /s

change in kinetic energy = m(30)^2/2 - m(-30)^2/2 = 0

if puck hits the net

velocity of pluck after collison = 0m/s due to inelastic collision

final momentum = 0.2* 0 = 0 kg m/s

change in momentum = 6 -(0) = 6kg m /s

change in kinetic energy = m(30)^2/2 - m(0)^2/2 = 90 J

thank you