A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s it strikes a second hockey puck with a mass 0.11 kg the first hockey puck comes to rest after the collision of conservation of momentum holds what is the total final momentum of the system (round your answer to the nearest tenth

Answers

Answer 1

Answer:

Conservation of momentum means that the final momentum must be the same as the initial momentum, where the momentum is defined as:

P = m*v

where m is mass, and v is velocity.

First, let's calculate Pi, the initial momentum.

Pi = Pi₁ + Pi₂

Where the subscripts 1 and 2 refer to each puck:

Pi₁ = 0.17kg*15m/s = 2.55 (kg*m/s)

Pi₂ = 0.11kg*0m/s = 0 (kg*m/s)

Pi = 2.55 (kg*m/s) + 0 (kg*m/s) = 2.55 (kg*m/s)

Now, after the collision we have that:

The first puck comes to rest, then:

Pf₁ = 0.17kg*0m/s =  0 (kg*m/s)

And the second puck will have a velocity v, that we want to find:

Pf₂ = 0.11kg*v

Pf = Pf₁ + Pf₂ = 0 (kg*m/s) +  0.11kg*v =  0.11kg*v

And by conservation of momentum we have:

Pi = Pf

2.55 (kg*m/s) = 0.11kg*v

We divide in both sides by 0.11kg

2.55 (kg*m/s)/0.11kg = v

23.2 m/s = v

And from this, the final momentum will be the same as the initial:

Pf = 23.2m/s*0.11kg = 2.55 (kg*m/s)

Answer 2

The total final momentum of the system is 2.55 Kg m/sec and this can be determined by conserving the momentum.

Given :

A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s it strikes a second hockey puck with a mass of 0.11 kg the first hockey puck comes to rest after the collision of conservation of momentum holds.

The formula of the momentum is given below:

P = mv

Now, let conserve the moment, that is:

[tex]\rm P_i=P_f[/tex]   --- (1)

where [tex]\rm P_i[/tex] is the initial momentum and [tex]\rm P_f[/tex] is the final momentum.

First, determine the initial momentum.

[tex]\rm P_i = 0.17\times 15 + 0.11\times 0= 2.55\;Kg\;m/sec[/tex]

Now, determine the final momentum.

[tex]\rm P_f = 0.17\times 0+ 0.11\times v = 0.11v\;Kg\;m/sec[/tex]

Now. substitute the value of [tex]\rm P_i[/tex] and [tex]\rm P_f[/tex] in the equation (1).

[tex]\rm 2.55=0.11v[/tex]

v = 23.18 m/sec

So, the final momentum can be calculated as:

[tex]\rm P = 23.18 \times 0.11[/tex]

P = 2.55 Kg m/sec

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Answers

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