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kodGreya [7K]
3 years ago
14

an object has a momentum of 250 kg m/s what would be the momentum of an object that has half of the mass and going at the same v

elovcity
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

The momentum of the second object is 125 kg m/s

Explanation:

The momentum of an object is given by

p=mv

where

m is the mass of the object

v is its velocity

For the object in this problem,

p = 250 kg m/s

And its mass is m and its velocity is v.

The second object has a mass of m' = \frac{m}{2} and same velocity v, so its momentum is

p'=m'v = (\frac{m}{2})v=\frac{1}{2}(mv)=\frac{1}{2}p

So, the second object has a momentum that is half of the momentum of the first object, therefore it is:

p'=\frac{1}{2}(250)=125 kg m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 6,000 kg is the mass of the first train

u_1 = 10 m/s is the initial velocity of the first train

m_2 = 4,000 kg is the mass of the second train

u_2 = 0 is the initial velocity of the second train  (initially at rest)

v is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s

2)

There are two types of collisions:

  • Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
  • Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J

After:

K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

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Answer:

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