1). the product of the two masses being gravitationally attracted to each other
2). the distance between their centers of mass
And that's IT. The gravitational force between them depends on
only those two things, nothing else.
Given that
Velocity of missile (v) = 20 m/s ,
Angle of missile (Θ) = 53°
Determine , Vertical component = v sin Θ
= 20 sin 53°
= 15.97 m/s
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
Answer:
Speed; v = 17 m/s
Explanation:
We are given;
Radius; r = 110m
Angle; θ = 15°
Now, we know that in circular motion,
v² = rg•tanθ
Thus,
v = √(rg•tanθ)
Where,
v is velocity
r is radius
g is acceleration due to gravity
θ is the angle
Thus,
v = √(rg•tanθ) = √(110 x 9.8•tan15)
v = √(288.85)
v = 17 m/s