A67.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 17.0 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 their final velocities (in m/s) be in this case? (assume the original direction of the ice puck toward the goalie is in the positive direction. indicate the direction with the sign of your answer.)

The final velocity of the puck is -16.9 m/s

The final velocity of the goalie is 0.0759 m/s

Explanation:

Hi there!

In an elastic collision, both the kinetic energy and the momentum of the system is conserved (i.e. the final and initial momentum and kinetic energy of the system is the same).

The initial momentum of the system is calculated as follows:

initial momentum =m1 · v1 + m2 · v2

Where:

m1 = mass of the goalie.

v1 = velocity of the goalie.

m2 = mass of the puck.

v2 = velocity of the puck.

initial momentum = 67.0 kg · 0 m/s + 0.150 kg · 17.0 m/s

initial momentum = 2.55 kg · m/s

The initial kinetic energy (KE) of the system is calculated in a similar way:

initial KE = 1/2 · m1 · v1² + 1/2 · m2 · v2²

initial KE = 1/2 · 67.0 kg · (0 m/s)² + 1/2 · 0.150 kg · (17.0 m/s)²

initial KE = 21.675 J

Now let´s express the final momentum of the system:

final momentum = m1 · v1´ + m2 · v2´

Where v1´ and v2´are the final velocities of the goalie and the puck respectively.

final momentum = 67.0 kg · v1´ - 0.150 kg · v2´

since initial momentum = final momentum, then:

2.55 kg · m/s = 67.0 kg · v1´ + 0.150 kg · v2´

Solving for v1´:

(2.55 kg · m/s - 0.150 kg · v2´) / 67 kg = v1´

In the same way, we can express the final KE of the system as follows:

final KE = 1/2 · m1 · (v1´)² + 1/2 · m2 · (v2´)²

since initial KE = final KE:

21.675 J = 1/2 · m1 · (v1´)² + 1/2 · m2 · (v2´)²

21.675 J = 1/2 · 67.0 kg  · (v1´)² + 1/2 · 0.150 kg · (v2´)²  

Let´s multiply by 2 both sides of the equation:

43.35 J = 67.0 kg  · (v1´)² + 0.150 kg · (v2´)²

Now, let´s replace v1´ with the expression obtained above (I will omit units for clarity):

43.35 = 67 (2.55  - 0.150 · v2´)²/ 67² + 0.150 · (v2´)²

43.35 =  (2.55  - 0.150 · v2´)² /67 + 0.150 · (v2´)²

Let´s multiply by 67 both sides of the equation:

2904.45  = (2.55  - 0.150 · v2´)²  + 10.05 · (v2´)²

2904.45  = 6.5025 - 0.765 v2´ + 0.0225 (v2´)² + 10.05 · (v2´)²

2904.45 = 6.5025 - 0.765 v2´ + 10.0725 (v2´)²

0 = -2904.45 + 6.5025 - 0.765 v2´ + 10.0725 (v2´)²

0 = -2897.9475 - 0.765 v2´ + 10.0725 (v2´)²

Solving the quadratic equation:

v2´ = -16.9 m/s  (the other value is 17 m/s which is discarded because it is the initial velocity).

The final velocity of the puck is -16.9 m/s

Now, let´s find the final velocity of the goalie:

(2.55 kg · m/s - 0.150 kg · v2´) / 67 kg = v1´

(2.55 kg · m/s - 0.150 kg · (-16.9 m/s)) / 67 kg = v1´

v1´ = 0.0759 m/s

The final velocity of the goalie is 0.0759 m/s

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

this could be newton#'s law of cooling. the cooling curve is exponential