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Dan is gliding on his skateboard at 4.0 m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.0 m/s. How fast is Dan going as his feet hit the ground? Dan’s mass is 50 kg and the skateboard’s mass is 5.0 kg?

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asuwafohjames

Answer:

3.6 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before jump = Total momentum after jump

Note: Before Dan jump off the skateboard, they where both moving with the same velocity

u(m+m') = mv+m'v'................. Equation 1

Where m = Dan's mass, m' = mass of the skateboard, u = common velocity before the jump, v = Dan's final velocity, v' = The final velocity of the skateboard.

make v the subject of the equation

v = [u(m+m')-m'v')]/m.............. Equation 2

Given: u = 4.0 m/s, m = 50 kg, m' = 5 kg, v' = 8 m/s

Substitute into equation 2

v = [4(50+5)-(5×8)]/50

v = (220-40)/50

v = 180/50

v = 3.6 m/s

Cricetus

The velocity will be "3.6 m/s".

Given:

  • u = 4.0 m/s
  • m = 50 kg
  • m' = 5 kg
  • v' = 8 m/s

By using Law of conservation of momentum,

→ [tex]u(m+m') = mv+m'v'[/tex]

or,

→ [tex]v = \frac{[u(m+m')-m'v']}{m}[/tex]

By substituting the values,

      [tex]= \frac{[4(50+5)-(5\times 8)]}{50}[/tex]

      [tex]= \frac{220-40}{50}[/tex]

      [tex]= \frac{180}{50}[/tex]  

      [tex]= 3.6 \ m/s[/tex]

Thus the above answer is appropriate.

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https://brainly.com/question/15841866

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