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A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the velocity of the skater?
a. -0.07 m/s
b. 0.15 m/s
c. 0.30 m/s
d. 0.07 m/s

Respuesta :

Khoso123

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]

So Option d is correct one

The velocity of skater is [tex]0.07m/s[/tex]

Option d is correct.

Law of conservation of momentum:

For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.

  • The expression is given as,

                     [tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

  • Given that, mass of  ice skater [tex]m_{1}=65kg[/tex], mass of snow ball [tex]m_{2}=0.15kg[/tex]
  •   Velocity of snow ball [tex]v_{2}=32m/s[/tex]
  • Substitute values in above expression.

                 [tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]

The velocity of skater is [tex]0.07m/s[/tex]

Learn more about the Velocity here:

https://brainly.com/question/24445340

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