Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col
2 answers:
In an inelastic collision, only momentum is conserved, while energy is not conserved.
1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find
, the velocity of the nail and the block after the collision:
2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
Rearranging the formula, we can find
, the velocity of the nail before the collision:
1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find
2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
Rearranging the formula, we can find
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Answer:
The value is
Explanation:
From the question we are told that
The initial velocity of the proton is
At a distance R from the nucleus the velocity is
The velocity considered is
Generally considering from initial position to a position of distance R from the nucleus
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
=>
=>
=>
And is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
=>
=>
Generally considering from initial position to a position of distance from the nucleus
Here represented the distance of the proton from the nucleus where the velocity is
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
=>
=>
=>
And is the change in electric potential energy from initial position to a position of distance
from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
=>
=>
Divide equation 2 by equation 1
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=>
=>
As we know that KE and PE is same at a given position
so we will have as a function of position given as
also the PE is given as function of position as
now it is given that
KE = PE
now we will have
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
now to find the fraction of kinetic energy
now since total energy is sum of KE and PE
so fraction of PE at the same position will be