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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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The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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