14J ÷ 8.5 = 1.64705882
Remember, 1.64705882 = v², so we need to find the square root.
The square root of 1.64705882 is 1.283377894464448
Hope this helps!
Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello
w disk is initially at rest and is struck by the orange disk moving with a speed of 4.60 m/s. After the collision, the orange disk moves along a direction that makes an angle of 38.0° with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.
1 answer:
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
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14J ÷ 8.5 = 1.64705882
Remember, 1.64705882 = v², so we need to find the square root.
The square root of 1.64705882 is 1.283377894464448
Hope this helps!
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
