What is the correct answer?
2 answers:
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Answer:
Explanation:
Given
Position of a car is given by
and angular speed is
angular acceleration is
at
Answer:
A. 1.4 m/s to the left
Explanation:
To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:
where:
M = momentum [kg*m/s]
M = m*v
where:
m = mass [kg]
v = velocity [m/s]
where:
m1 = mass of the basketball = 0.5 [kg]
v1 = velocity of the basketball before the collision = 5 [m/s]
m2 = mass of the tennis ball = 0.05 [kg]
v2 = velocity of the tennis ball before the collision = - 30 [m/s]
v3 = velocity of the basketball after the collision [m/s]
v4 = velocity of the tennis ball after the collision = 34 [m/s]
Now replacing and solving:
(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)
1 - (0.05*34) = 0.5*v3
- 0.7 = 0.5*v
v = - 1.4 [m/s]
The negative sign means that the movement is towards left
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