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musickatia [10]
3 years ago
10

A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is

initially at rest. After the collision the green block is at rest and the red block is moving to the right.
How does m compare to M ?
Physics
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

M is equal to m

Explanation:

In case we say that the green block's mass m is less than red block's mass M, then the green block would have bounced and moved back to the left instead of coming to rest. The other case where if mass of green block's mass m would have been greater than the red block's mass M, the green block would have kept moving to the right instead of coming to rest. After collision, the red block moves to the right because of exchange of velocities. Therefore, m=M since m comes to rest and M moves to the right

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂,  which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

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a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf 

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
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b) Here use the equation vf^2=vi^2+2gd 

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so in here we are looking for the vi so let us isolate that variable 
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therefore,</span></span>

vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s

<span>c) The energy was converted to heat due to friction with the air and the ground.</span></span>

6 0
3 years ago
IM SOO CONFUSED PLS HELP!! The mass of the nucleus is approximately EQUAL to the mass number multiplied by ____ Atomic Mass unit
nevsk [136]

Answer:

option a.

Explanation:

We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.

So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.

Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u

Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u

(where u = atomic mass unit)

Then the weight of the nucleus is about A times 1u, or:

A*1u = A atomic mass units.

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The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.

option a.

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Second and third class levers are differentiated by __________. a. the location of the fulcrum b. the location of the load c. th
strojnjashka [21]

Answer:

the answer is b

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Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>

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Answer:

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2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

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