Define velocity ratio in pulley
1 answer:
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Answer:
6.25 m/s
Explanation:
mass of man (m1) = 80 kg
mass of boy (m2) = 20 kg
mass of man and boy after collision (m12)= 20 + 80 = 100 kg
velocity of man and boy after collision (v) = 2.5 m/s
angle θ = 60 °
How fast was the boy moving just before the collision ?
- From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
- M₁₂ = total momentum after collision = m12 x v = 100 x 2.5 = 250
- Mboy = momentum of the boy before collision = m2 x Velocity of boy
- Mman = momentum of the man before collision = m1 x velocity of man
- from the triangle, cos θ =
cos 60 =
Mboy = 250 x cos 60 = 125
- recall that momentum of the boy (Mboy) also = m2 x Velocity of boy
therefore
125 = 20 x velocity of boy
velocity of boy = 125 / 20 = 6.25 m/s
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, ................ii
form i and ii we can write
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v