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A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg and is at rest. Assuming a closed system, find the total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper. In 3-4 sentences, identify the object with the greater momentum after the puck is caught and explain your reasoning.

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abidemiokin

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

Answer: the goalkeeper has the greater momentum after the puck was caught.

Explanation:

Mass of puck - 0.16 kg

Velocity of puck - 40 m/s

Mass of goalkeeper - 120 kg

Velocity of goalkeeper (at rest) - 0 m/s

m1v1 + m2v2 (total momentum of goalkeeper and puck)

0.16(40) + 120(0)

6.4 + 0

6.4 kg m/s

(Velocity) m1u1 + m2u2 = (m1 + m2)v

6.4 = (0.16 + 120)v

6.4 = 120.16v

v = 0.053 m/s

m2v = momentum of the goalkeeper after the collision = 120(0.053) = 6.36 kg m/s

First, I found the total momentum of the goalkeeper and the puck which is 6.4 kg m/s. Next, to find the velocity, I equaled the total momentum of the goalkeeper and the puck to the mass of the puck and the mass of the goalkeeper and added them together. Then, I divided the total momentum of the goalkeeper and the puck by the mass of the puck and the mass of the goalkeeper added together and got the velocity which is 0.053 m/s. Once I found the total momentum of the goalkeeper to get the velocity, I found the momentum of the goalkeeper after the collision by multiplying the mass of the goalkeeper by the velocity and got 6.36 kg m/s. Since the momentum of the goalkeeper after the collision is greater than the puck, the goalkeeper has the greater momentum after the puck was caught.

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