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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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Schach [20]

Answer:

q  =  -461532.5 \ C

Explanation:

From the question we are told that

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Generally according to Gauss law

=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

          A is the surface  area of the of the earth which is mathematically represented as

     A  =  4\pi r^2

Where r is the radius of the earth which has a value r = 6.3781*10^6 m

 substituting values

    A  = 4 * 3.142  *   (6.3781*10^6 \ m)^2

    A  =5.1128 *10^{14} \ m^2

So

   q  =  -E  * A  *  \epsilon _o

Here \epsilon_o s the permitivity of free space with value

          \epsilon_o  =  8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

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6 0
3 years ago
If something is traveling at 20 m/s constant velocity, is it in equilibrium? If a projectile is launched at a velocity of 20 m/s
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If a real projectile is launched, the force of gravity acts on it vertically downward.  There's no upward force acting on it to balance gravity.  Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium.  Too bad.


3 0
2 years ago
If you make multiple measurements of your height, you are likely to find that the results vary by nearly half an inch in either
Lisa [10]

Answer:

Height h= 1.7 m

Explanation:

Supposing we have to find height in meter.

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Given that:

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Therefore total height of a man in meter

5 feet 7 inch = 1.5424+0.1778 =1.7 m

Height h= 1.7 m

8 0
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