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Answer:
When the person is drunk the car travels 20.1 meters more than when the person is sober. The distance traveled by car is:
Person drunk: 28.1 m
Person sober: 8 m
Explanation:
The distance traveled can be found using the following equation:
Where:
v: is the speed = 90 km/h
t: is the time
When the person is sober, the time that takes to hit the brakes is 0.320 s, so we have:
And when the person is drunk, the time is 1.00 s, hence the distance is:
The distance traveled by the car when the person is drunk compared when the person is sober is:
Therefore, when the person is drunk the car travels 20.1 meters more than when the person is sober.
I hope it helps you!
Answer:
Vf₂ = 0.29 m/s :Speed of the block immediately after the collision.
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction and direction as 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 : Initial linear momentum quantity
Nomenclature and data
m₁: ball mass= 5.21 g= 5.21*10⁻³kg
V₀₁: initial ball speed, =412 m/s
Vf₁: final ball speed
m₂: block mass = 14.8 kg
V₀₂: initial block speed, = 0
Vf₂: final block speed
Problem development
For this problem the collision is perfectly elastic ,then, In addition to the linear moment, the kinetic energy is also conserved.
We assume that the ball moves to the right before the collision y continues moving to the right after the collision .(+)
We assume that the block moves to the right before the collision.(+)
We apply furmula (1)
P₀=Pf
m₁*V₀₁+m₂*V₀₂=m₁*Vf₁+m₂*Vf₂
5.21*10⁻³*412+14.8*0= 5.21*10⁻³*Vf₁+14.8*Vf₂
2.15= 5.21*10⁻³*Vf₁+14.8*Vf₂ Equation (1)
For perfectly elastic collision the coefficient of elastic restitution (e) is equal to 1, and e is defined like this:
1*(V₀₁-V₀₂) =Vf₂-Vf₁ , V₀₂=0, V₀₁ =412 m/s
412=Vf₂-Vf₁
Vf₁=Vf₂-412 Equation (2)
We replace Equation (2) in Equation (1)
2.15= 5.21*10⁻³(Vf₂-412)+14.8*Vf₂
2.15= 5.21*10⁻³*Vf₂-2.15+14.8*Vf₂
4.3=14.805Vf₂
Vf₂ =4.3/14.805= 0.29 m/s : (+) ,with equal direction of the movement of the ball before the collision.