Answer:
The puck B remains at the point of collision.
Explanation:
This is an elastic collision, so both momentum and energy are conserved.
The mass of both pucks is m.
The velocity of puck B before the collision is vb.
The velocity of puck A and B after the collision is va' and vb', respectively.
Momentum before = momentum after
m vb = m vb' + m va'
vb = vb' + va'
Energy before = energy after
½ m vb² = ½ m vb'² + ½ m va'²
vb² = vb'² + va'²
Substituting:
(vb' + va')² = vb'² + va'²
vb'² + 2 va' vb' + va'² = vb'² + va'²
2 va' vb' = 0
va' vb' = 0
We know that va' isn't 0, so:
vb' = 0
The puck B remains at the point of collision.
Answer: 10 m/s
We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.
Answer:
t = 0.24 s
Explanation:
As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:
Translation: ΣF = ma
Rotation: ΣM = Iα ; where α = angular acceleration
Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:
ΣM = I(a/R)
Now we are going to resolve and combine these equations.
For translation: Fx - Ffr = ma
We know that Fx = mgSin27°, so we substitute:
(1) mgSin27° - Ffr = ma
For rotation: (Ffr)(R) = (2/3mR²)(a/R)
The radius cancel each other:
(2) Ffr = 2/3 ma
We substitute equation (2) in equation (1):
mgSin27° - 2/3 ma = ma
mgSin27° = ma + 2/3 ma
The mass gets cancelled:
gSin27° = 5/3 a
a = (3/5)(gSin27°)
a = (3/5)(9.8 m/s²(Sin27°))
a = 2.67 m/s²
If we assume that the acceleration is a constant we can use the next equation to find the velocity:
V = √2ad; where d = 0.327m
V = √2(2.67 m/s²)(0.327m)
V = 1.32 m/s
Because V = d/t
t = d/V
t = 0.327m/1.32 m/s
t = 0.24 s
Answer:
The mass of the cargo is 
Explanation:
From the question we are told that
The radius of the spherical balloon is 
The mass of the balloon is 
The volume of the spherical balloon is mathematically represented as
substituting values
The total mass the balloon can lift is mathematically represented as
where 
is the density of helium with a value of
and 
is the density of air with a value of
substituting values
Now the mass of the cargo is mathematically evaluated as
The density of the material would be 4.1 g/cm³.
Density is calculated by dividing the mass by the volume.
D=m÷v
D=45 g÷11 cm³
D=4.1 g/cm³
