Respuesta :
Answer:
a. [tex]v'=8.925\ m.s^{-1}[/tex]
b. [tex]I=1585397.5872\ N.s[/tex]
c. [tex]s=265.4125\ m[/tex]
Explanation:
- mass of moving car, [tex]m_m=2100\ kg[/tex]
- velocity of moving car, [tex]v=17\ m.s^{-1}[/tex]
- mass of the stationary car, [tex]m_s=1900\ kg[/tex]
- coefficient of kinetic friction, [tex]\mu_k=0.68[/tex]
a.
Velocity of the two cars just after the collision:
Using the law of conservation of
[tex]m_m.v+m_s\times 0=(m_m+m_s)\times v'[/tex]
[tex]2100\times 17+ 0=(2100+1900)\times v'[/tex]
[tex]v'=8.925\ m.s^{-1}[/tex]
b.
Frictional force acting on the combined mass of cars:
[tex]f=\mu_k\times N[/tex]
where:
N = reaction normal to the contact surface by the ground
[tex]f=0.68\times (2100+1900)\times 9.8[/tex]
[tex]f=26656\ N[/tex] is the resistance force to the motion of the wheels
Now, by Newton's second law of motion:
[tex]f=\frac{dp}{t}[/tex]
[tex]f=\frac{(m_m+m_s)\times v'}{t}[/tex]
[tex]26656=\frac{(2100+1900)\times 8.925}{t}[/tex]
[tex]t=59.4762\ s[/tex]
Now impulse is the impact load acting acting for certain time:
[tex]I=f\times t[/tex]
[tex]I=26656\times 59.4762[/tex]
[tex]I=1585397.5872\ N.s[/tex]
c.
The distance the the cars skid before coming to rest:
using equation of motion:
[tex]v'_f=v'+a.t[/tex]
we've [tex]v'_f=0\ m.s^{-1}[/tex] since final velocity is zero
[tex]0=8.925+a\times 59.4762[/tex]
[tex]a=0.1501\ m.s^{-2}[/tex]
now since we have:
[tex]s=v'.t+\frac{1}{2} a.t^2[/tex]
[tex]s=8.925\times 59.4762-0.5\times 0.1501\times 59.4762^2[/tex]
[tex]s=265.4125\ m[/tex]
A. The velocity of the two cars just after the collision is 8.925 m/s
B. The impulse that acts on the skidding cars just after the collision is 35700 Ns
C. The distance travelled by the skidding cars before coming to rest is 5.98 m
A. Determination of the velocity the two cars
Mass of moving car (m₁) = 2100 Kg
Initial velocity of moving car (u₁) = 17 m/s
Mass of stationary car (m₂) = 1900 Kg
Initial velocity of stationary car (u₂) = 0 m/s
Velocity of the two cars (v) =?
[tex]v = \frac{u_1m_1 + u_2m_2}{m_1m_2} \\ \\ v = \frac{(2100 \times 17) + (1900 \times 0)}{2100 + 1900} \\ \\ v = 8.925 \: m/s[/tex]
Thus, the velocity of the two cars is 8.925 m/s
B. Determination of the impulse
Mass of moving car (m₁) = 2100 Kg
Mass of stationary car (m₂) = 1900 Kg
Velocity of the two cars (v) = 8.925 m/s
Impulse (I) =?
I = Ft = mv
I = (2100 + 1900) × 8.925
I = 35700 Ns
Thus, the impulse on the skidding cars is 35700 Ns
C. Determination of the distance travelled by the skidding cars.
- We'll begin by calculating the force acting on the skidding cars
Mass of moving car (m₁) = 2100 Kg
Mass of stationary car (m₂) = 1900 Kg
Total mass = 2100 + 1900 = 4000 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (N) = mg = 4000 × 9.8 = 39200 N
Coefficient of kinetic friction (μ) = 0.68
Frictional force (F) =?
F = μN
F = 0.68 × 39200
F = 26656 N
- Next, we shall determine the time.
Impulse (I) = 35700 Ns
Force (F) = 26656 N
Time (t) =?
I = Ft
35700 = 26656 × t
Divide both side by 26656
t = 35700 / 26656
t = 1.34 s
- Finally, we shall determine the distance travelled
Initial velocity of the two cars (u) = 8.925 m/s
Final velocity of the two cars (v) = 0 m/s
Time (t) = 1.34 s
Distance (s) =?
[tex]s = \frac{(v + u)t}{2} \\ \\ s = \frac{(8.925 + 0)1.34}{2} \\ \\ s = \frac{8.925 \times 1.34}{2} \\ \\ s = 5.98 \: m[/tex]
Therefore, the distance travelled by the skidding cars before coming to rest is 5.98 m
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