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Gnoma [55]
3 years ago
14

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

stop but imparts all it’s mometum to the truck. How fast is the truck moving after the collision and in which direction? (Hint: this is an elastic collision)
m1v1 + m2v2 = m1v1 + m2v1
Physics
2 answers:
Burka [1]3 years ago
8 0

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

ioda3 years ago
4 0
<h2>Answer:</h2>

5m/s to the south

<h2>Explanation:</h2>

Using the principle of conservation of linear momentum which states that the total momentum of colliding bodies before collision is equal to the total momentum of these bodies after collision. In other words, given two bodies of masses m₁ and m₂ and initial velocities u₁ and u₂ respectively. When these two bodies collide and they begin to move with final velocities of v₁ and v₂ respectively, the momentum before they collide is equal to the momentum after they collide. i.e

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂        ------------------(i)

From the question, the bodies are a car and a truck.

Therefore;

m₁ = mass of car = 1500kg

m₂ = mass of the truck = 4500kg

u₁ = initial velocity of the car = +15.0m/s (taking due south as positive)

u₂ = initial velocity of the truck = 0m/s  (since the tuck is at rest)

v₁ = final velocity of the car = 0m/s (since the car comes to a stop)

v₂ = final velocity of the truck

Substitute these values into equation (i) as follows;

1500(15.0) + 4500(0) = 1500(0) + 4500(v₂)

22500 + 0 = 0 + 4500v₂

22500 = 4500v₂

Solve for v₂;

v₂ = \frac{22500}{4500}

v₂ = +5m/s

Therefore, the velocity at which the truck is moving is 5m/s and since this value is positive and south is taken as positive, the truck is moving south.

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Given:

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  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
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<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

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<u>For plywood-Styrofoam interface from inside:</u>

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20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

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