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A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away

Respuesta :

Manetho

Answer:

0.0241 m

Explanation:

mass of the hockey player m1 = 90 kg

mass of puck m2 = 0.150 kg

puck velocity v1= 45 m/s

distance traveled by puck to reach the goal =15.0 m.

now accoding to momentum conservation law

90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]

therefore, v2= -0.0725 m/s.

Now time taken by the puck to reach the goal

t= 15/45 = 1/3 sec.

therefore, how far does the player recoil in the time

=0.0725×1/3= 0.0241 m.

nuhulawal20

the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

Given the data in the question

  • Mass of the player; [tex]m_1 = 90.0kg[/tex]
  • Mass of puck; [tex]m = 0.150kg[/tex]

Since they were both at rest initially

  • Initial velocity of player; [tex]u_1 = 0[/tex]
  • Initial velocity of puck; [tex]u = 0[/tex]
  • Velocity of player after the hit; [tex]v_1 = \ ?[/tex]
  • Velocity of puck after the hit; [tex]v = 45.0m/s[/tex]
  • Distance to the goal; [tex]s = 15.0m[/tex]

Using conservation of liner momentum:

[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]

Now, Since they were both at rest initially

[tex]0 = mv+ m_1v_1[/tex]

We substitute in our values to find the velocity of the player after the hit ( recoil velocity )

[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]

{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }

Hence, recoil velocity of the player is 0.075m/s

Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute our values into the expression

[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]

Hence, the time taken for the puck to reach the goal is 0.3333 seconds.

Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute in our values

[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]

Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

Learn more: https://brainly.com/question/3637213

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