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Vesna [10]
3 years ago
15

A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

What is the final velocity of the two combined vehicles?
Physics
1 answer:
IgorLugansk [536]3 years ago
7 0

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]

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A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
SpyIntel [72]
This is a problem of conservation of momentum

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A) man throws the rock forward

=>

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V1 = 14.5 m/s, in the same direction of the sled with the man

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Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

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Which of the following is a field (action-at-a-distance) force?
Kobotan [32]

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Explanation:

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What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
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We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

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  • a is acceleration

In our case,

  • F = ?
  • m = 2500 kg
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\tt \: F_{net}  = 2500 \times 20 \\   \tt= 50000

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

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