A skier moving at 5.23 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
1 answer:
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Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
Answer:
b) = 4.88 cm
, c)
’= 1/M (m₁ d₁ + m₃ d₃) and d)
’= 1.88 cm
Explanation:
The definition of mass center is
= 1/M ∑ xi mi
Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.
Part b
Apply this equation to our case.
Body 1
They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left
Body 2
They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)
D₂ = d₁ + d₂
D₂ = 1.1 + 1.9
D₂ = 3.0 cm
Body 3
Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin
D₃ = D₂ + d₂
D₃ = 3.0 + 3.9
D₃ = 6.9 cm
With this data we substitute and calculate the center of mass
M = m₁ + m₂ + m₃
M = 24 + 12 + 56
M = 92 g
= 1/92 (1.1 24 + 3.0 12 + 6.9 56)
= 1/92 (448.8)
= 4,878 cm
= 4.88 cm
This distance is from the left end of the bar
Par c)
In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.
Body 1
It is at d1 = -1.9 cm
It is negative for being on the left and the value is the relative distance of 1 to 2
Body 2
d2 = 0 cm
The reference system for her
Body 3
d3 = 3.9 cm
Positive because that is to the left of the reference system and is the relative distance between 2 and 3
Let's write the new center of mass (xcm')
’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)
’= 1/M (m₁ d₁ + m₃ d₃)
Part d) Let's calculate the value of the center of mass
’= 1/92 ((24 (-1.9) +56 3.9)
’= 1/92 (172.8)
’= 1.88 cm
This distance is to the right of the central marble