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jenyasd209 [6]
2 years ago
12

A 5000-kg freight car collides with a 10,000-kg freight car at rest. They couple upon collision and move at 2 m/s. What was the

initial speed of the 5000-kg car? 4 m/s 5 m/s 6 m/s 8 m/s none of these.
Physics
1 answer:
Ann [662]2 years ago
6 0

The initial speed of the 5000-kg car is 6 m/s.

To calculate the initial speed of the car, we apply the law of conservation of momentum

<h3>Law of conservation of momentum</h3>

The Law of conservation of momentum states that for two colliding bodies, the total momentum of the system before the collision is equal to the total momentum after collision provided there is no net external force acting on the system.

Formula:

  • mu+m'u' = V(m+m')............... Equation 1

Where:

  • m = mass of the first car
  • m' = mass of the second car
  • u = initial speed of the first car
  • u' = initial speed of the second car
  • V = common speed.

From the question,

Given:

  • m = 5000 kg
  • m' = 10000 kg
  • u' = 0 m/s (at rest)
  • V = 2 m/s

Substitute these values into equation 1

  • 5000(u)+10000(0) = 2(5000+10000)
  • 5000u+0 = 30000
  • u = 30000/5000
  • u = 6 m/s.

Hence, The initial speed of the 5000-kg car is 6 m/s.

Learn more about the law of conservation of momentum here: brainly.com/question/7538238

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slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa
xenn [34]

Answer:

Maximum height reached by the rocket is

y_{max} = 308 m

total time of the motion of rocket is given as

T = 16.44 s

Explanation:

Initial speed of the rocket is given as

v_i = 50 m/s

acceleration of the rocket is given as

a = 2 m/s^2

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 50^2 = 2(2)(150)

v_f = 55.68 m/s

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 55.68^2 = 2(-9.81)(y - 150)

y_{max} = 308 m

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

v_f - v_i = at

55.68 - 50 = 2 t

t_1 = 2.84 s

2) time to reach ground from this height

\Delta y = v_y t + \frac{1}{2}gt^2

-150 = 55.68 t - \frac{1}{2}(9.81) t^2

t_2 = 13.6 s

so total time of the motion of rocket is given as

T = 13.6 + 2.84 = 16.44 s

3 0
3 years ago
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