A65-kg ice hockey goalie, originally at rest, catches a 0.145-kg hockey puck slapped at him at a velocity of 35 m/s. suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. what would the final velocities of the goalie and the puck be in this case? assume that the collision is completely elastic. v goalie = m / sv puck = m / s

The velocity of the goalie = 0.156 m/s

The velocity of the puck is 34.85 m/s, but in the reverse direction (-34.85 m/s)

Explanation:

The total momentum before equals the totam momentum after the collision for objects with mass m1 and m2  

m1*vi1 + m2 vi2 = m1 vf1 + m2 vf2

⇒ with vi = the initial velocity

⇒ with vf = the final velocity

initial momentum = final momentum  

An elastic collision is one in which the total kinetic energy of the two colliding objects is the same before and after the collision. For an elastic collision, kinetic energy is conserved. That is:  

0.5 m1 vi1² + 0.5 m2 vi2² = 0.5 m1 vf1² + 0.5 m2 vf2²  

Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:  

vf1 = [(m1 - m2) vi1 + 2m2 vi2]/[m1 + m2]  

vf2 = [2m1vi1 − (m1 - m2) vi2]/[m1 + m2]  

⇒ with m1 = the mass of the goalie = 65 kg

⇒ with m2 = the mass of the puck = 0.145 kg

⇒ with vi1 = the initial velocity of the goalie 0 m/s

⇒ with vi2 = the initial velocity of the puck = 35 m/s

⇒ with vf1 = The final velocity of the goalie

   => vf1 = [(65 - 0.145)0 + 2(0.145)35]/[65.145]  

=  0.1558  ≈ 0.156 m/s. Since it's positve this is in the direction of the puck

=> vf2 = [0 − (65 - 0.145)35]/[65.145] = -34.85 m/s

The velocity of the puck is 34.85 m/s, in the reverse direction. (Has a negative sign)