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Yuri [45]
3 years ago
11

Two ice skaters are initially at rest. They push off and move in opposite directions. Ice Skater 1 has a mass of 90 kg and a vel

ocity of -10 meters per second. Ice Skater 2 has a mass of 105 kg. What is his velocity?
Physics
1 answer:
makvit [3.9K]3 years ago
4 0

Answer:

vB = 8.57[m/s]

Explanation:

Since there is no external force that generates momentum, the amount of momentum is conserved so you will have the following equation

m_{A}*v_{A}=m_{B}*v_{B}\\Where:\\m_{A}= 90[kg]\\v_{A}=10[m/s]\\m_{B}=105[kg]\\\\Therefore:\\\\v_{B}=\frac{m_{A}*v_{A}}{m_{B}}\\ v_{B}=\frac{90*10}{105}\\v_{B}=8.57[m/s]

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In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. I
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Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, m₂ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, v₂₂ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ × v₁₁ + m₂ × v₁₂ = m₁ × v₂₁ + m₂ × v₂₂

Substituting the known values;

65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

∴ 65 × v₂₁ + 45 × 1.5 = 0

45 × 1.5 = - 65 × v₂₁

v₂₁ = 45 × 1.5/(-65) ≈ -1.04 m/s

The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

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Q1. In a 100m track event, the time taken by four runners is 10 sec., 10.2 sec., 10.4 sec and 10.6 sec. Find the ratio of their
Anna35 [415]
Distance (s)=100m
time
T₁=10
T₂=10.2
T₃=10.4
T₄=10.6
by v=\frac{s}{t}
we get
V₁=\frac{100}{10}
V₁=10m/s

V₂=\frac{100}{10.2}
V₂=9.8m/s

V₃=\frac{100}{10.4}
V₃=9.61 m/s

V₄=\frac{100}{10.6}
V₄=9.43 m/s

V₁:V₂:V₃:V₄ = 10 : 9.8 : 9.61 : 9.43
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