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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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3 years ago
A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

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2 years ago
Which statement about the properties of matter is true? *25*
mojhsa [17]

Answer:

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3 0
3 years ago
The windshield of a car has a total length of arm and blade of 9 ​inches, and rotates back and forth through an angle of 93degre
Sergio039 [100]

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7​-in wiper​ blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

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