A baseball leaves a bat with a horizontal velocity of 20.0 m/s. When it has left the bat for a time of 0.250 s (Assume air resis
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Going down the first ramp:
• net force parallel to the ramp:
∑ <em>F</em> = <em>W</em> sin(60°) = <em>m</em> <em>a</em>₁
(<em>W</em> for <u>w</u>eight)
• net force perpendicular to the ramp:
∑ <em>F</em> = <em>N</em> + <em>W</em> cos(60°) = 0
(<em>N</em> for <u>n</u>ormal force)
The body has mass 0.1 kg, and with <em>g</em> = 10 m/s², its weight is <em>W</em> = 1 N. So in the first equation, we get
(1 N) sin(60°) = (0.1 kg) <em>a</em>₁ → <em>a</em>₁ ≈ 8.7 m/s²
Let <em>d</em>₁ be the distance the body moves down the ramp, i.e. the distance along the ramp between the starting point and the point <em>O</em>. Then
sin(60°) = <em>h</em> / <em>d</em>₁ → <em>d</em>₁ = 2<em>h</em>/√(3) ≈ 1.15<em>h</em>
Given an initial speed <em>v</em>₁ = 3 m/s, we find the speed <em>v</em>₂ at point <em>O</em> to be
<em>v</em>₂² - (3 m/s)² = 2 (8.7 m/s²) <em>d</em>₁
<em>v</em>₂ ≈ √(9 m²/s² + (17.4 m/s²) (1.15<em>h</em>))
<em>v</em>₂ ≈ √(9 m²/s² + (20 m/s²)<em> h</em>)
Going up the second ramp:
• net parallel force:
∑ <em>F</em> = -<em>Fr</em> - <em>W</em> sin(30°) = <em>m</em> <em>a</em>₂
(<em>Fr</em> for <u>fr</u>iction)
• net perpendicular force:
∑ <em>F</em> = <em>N</em> - <em>W</em> cos(30°) = 0
sin(30°) = cos(60°), and cos(30°) = sin(60°), so the second equation gives <em>N</em> = 0.87 N. Then with <em>µ</em> = 0.1, we have <em>Fr</em> = <em>µ</em> <em>N</em> = 0.087 N. The first equation then gives
-0.087 N - 0.5 N = (0.1 kg) <em>a</em>₂ → <em>a</em>₂ ≈ -5.9 m/s²
We now have
tan(30°) = <em>h</em>/<em>R</em> → <em>h</em> = (2.5 m)/√3 ≈ 1.4 m
(which, by the way, tells us that <em>v</em>₂ ≈ 6.2 m/s)
Then the distance traveled up the ramp is
<em>d</em>₂ = √(<em>h</em>² + <em>R</em> ²) ≈ 2.9 m
Use this to solve for the speed at the top of the ramp:
<em>v</em>₃² - <em>v</em>₂² = 2 (-5.9 m/s²) <em>d</em>₂
<em>v</em>₃ = √((6.2 m/s)² - (11.8 m/s²) (2.9 m)) ≈ 2.0 m/s
At the top of the second ramp:
• net parallel force:
∑ <em>F</em> = -<em>Fsp</em> - <em>W</em> sin(30°) = <em>m a</em>₂
(<em>Fsp</em> for <u>sp</u>ring)
• net perpendicular force:
∑ <em>F</em> = <em>N</em> - <em>W</em> cos(30°) = 0
By Hooke's law, <em>Fsp</em> = <em>kx</em>, so in the first equation we get
-<em>k</em> (0.10 m) - (1 N) cos(60°) = (0.1 kg) (-5.9 m/s²)
→ <em>k</em> ≈ 0.87 N/m
