The energy diagram shows the changes in energy during a chemical reaction.
1 answer:
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Answer:
Explanation:
An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is
p = mv and
p = (3520 + 1480)(13.6) so
p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is
which, in words, is
the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:
and
and
and
and
so
v = 13.3 m/s at 72.6°
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here,
Here,
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.