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borishaifa [10]
3 years ago
14

Why do we add the masses together after that inelastic collision?

Physics
1 answer:
fomenos3 years ago
5 0

Answer:An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.

hope this helps have a nice day❤️

Explanation:

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A weightlifter deadlifts a 300 kg weight. If the weightlifter has a mass of 100 kg, what is the force acting on his legs?
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Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as
Dmitriy789 [7]

Answer:

a) v=5.6725\,m.s^{-1}

b) h= 1.6420\,m

Explanation:

Given:

  • mass of the body, M=20\,kg
  • mass of the tyre,m=10\,kg
  • length of hanging of tyre, l=3.5m
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(a)

Using the equation of motion :

v^2=u^2+2a.d..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

u=0m.s^{-1}

putting the values in eq. (1)

v^2=0^2+2\times 3.62 \times 10

v=8.5088\,m.s^{-1}

Now, the momentum of the body just before the jump onto the tyre will be:

p=M.v

p=20\times 8.5088

p=170.1764\,kg.m.s^{-1}

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

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v'=5.6725\,m.s^{-1}

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

\frac{1}{2} (M+m).v'^2=(M+m).g.h

\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h

h\approx 1.6420\,m

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Answer:

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