Answer:
(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^
Explanation:
Given
W = 0.56N = mg
m = 0.56/g = 0.56/9.8 = 0.057kg
t = 3.00ms = 3.00×10-³s
Impulse is a vector quantity so we would treat it as such
We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that
u = (20.0m/s)ı^−(4.0m/s)ȷ^
ux = 20m/s
uy = – 4.0m/s
F = – (380N)ı^+(110N)ȷ^
Fx = –380N
Fy = 110N
J = impulse = force × time = F×t
So Jx = Fx ×t
Jy = Fy×t
Jx = –380×3×10-³ = -1.14Ns
Jy = 110×3×10-³ = 0.330Ns
Impulse also equals the change in momentum of the body. So
J = m(v–u)
J/m = v – u
V= J/m + u
Vx = Jx/m + ux
Vx = –1.14/0.057 + 20
Vx = -20 + 20 = 0m/s
Vx = 0m/s
Vy= Jy/m + uy
Vy= 0.33/0.057 + (-4.0)
Vy= 5.79 + (-4.0) = 1.79m/s
V = (0m/s)ı^−(1.79m/s)ȷ^
I THINK it is C. KEYWORD is THINK
Answer:
The x-coordinate of the particle is 24 m.
Explanation:
In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion
Xf=Xo+Voxt+0.5axt²(I)
Yf=Yo+Voyt+0.5ayt² (II)
Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.
The particle starts from rest from the origin, therefore:
Vox=Voy=0
Xo=Yo=0
Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:
12=0+(0)t+ 0.5(1.0)t²
12=0.5t²
Dividing by 0.5 and extracting thr squareroot both sides:
t=√12/0.5
t=√24 = 2√6
Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:
Xf=0+0t+0.5(2.0)(2√6)²
Xf= 24 m
