If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Answer:
vf₁ = 6.86 m/s , to the right
vf₂ = 2.96 m/s, to the right
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁= 0.220 kg : mass of object₁
m₂= 0.345 kg : mass of object₂
v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁
v₀₂= 6 m/s, to the right i :initial velocity of m₂
Problem development
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂
We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:
(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂
2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)
(2.1 - 6 ) = (vf₂ -vf₁)
-3.9 = (vf₂ -vf₁)
vf₂ = vf₁ - 3.9
vf₂ = vf₁ - 3.9 Equation (2)
We replace Equation (2) in the Equation (1)
2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)
2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)
2.532 + 1.3455 = (0.565)*vf₁
3.8775 = (0.565)*vf₁
vf₁ = (3.8775) / (0.565)
vf₁ = 6.86 m/s, to the right
We replace vf₁ = 6.86 m/s in the Equation (2)
vf₂ = 6.86 - 3.9
vf₂ = 2.96 m/s, to the right