could someone correct these for me? show work to help me understand thank you y
ID: 1584667 • Letter: C
Question
could someone correct these for me? show work to help me understand thank you
y 3/2). You are res, re. Use this study guide to he ain to use your textbook to f r, this is a GUIDE not a (30 0 points) A 25.0 kg PeeWee hockey goalie is standing motionless in front of the goal. The alie then catches a heavy 760 kg puck moving at 42.0 m/s. Assume that the contact between rictionless. the goalie's skates and the ice is entirely f KE PE (a) (6 points) What is the initial magnitude of the hockey puck's momentum 31,2 (b) (6 points) What is the final speed of the goalie after the puck is caught? rn 21 (c) (6 points) What is the momentum of the goalie after catching the puck? 7 (d) (6 points) If the catch takes 0.039 seconds, what is the average force exerted by the jot hockey puck on the goalie? 239 18. (e) (6 points) What happens to the kinetic energy of the system (both the hockey puck and the goalie make up the system) when the goalie cajolkes the puck? (Please cirele the REMAINES CONSTANT correct choice DECREASE INCREASESExplanation / Answer
(a) Initial magnitude of hockey puck's momentum -
Pi = m*Vi = 0.760 * 42.0 = 31.92 kg.m/s
(b) Apply conservation of momentum.
Pi = Pf
=> m*Vi = (M+m)*Vf
=> Vf = m*Vi / (M+m) = 31.92 / (25+0.76) = 1.24 m/s
(c) Momentum of the goalie after catching the puck = M*Vf = 25*1.24 = 31 kg.m/s
(d) Change in momentum of hockey puck = 31.92 - m*Vf = 31.92 - 0.76*1.24 = 30.98 kg.m/s
Therefore force exerted by it = Change in momentum / time = 30.98 / 0.039 = 794.4 N
(e) Initial kinetic energy Ei = (1/2)*m*Vi^2 = 0.5 * 0.76 * 42^2 = 670.32 J
Final kinetic energy of the system = 0.5*0.76*1.24^2 + 0.5*25*1.24^2 = 0.58 + 19.22 = 19.80 J
Therefore, the kinetic energy of the system will decrease.
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