Describe the process you used to build a model
1 answer:
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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call
the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
(1)
where
is the mass of ball A
is the mass of ball B
is the initial velocity of ball A
is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)
If we solve (1) to find, we find that the final velocity of the balls is
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
is the final velocity of ball A
is the final velocity of ball B
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
If we call
where
If we solve (1) to find
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
Answer:
greater speed will be obtained for the elastic collision,
Explanation:
To answer this exercise we must find the speed that the sail acquires after each impact.
Let's start by hitting a ball of clay.
The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.
initial instant. before the crash
p₀ = m v₀
where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest
final instant. After the crash
the mass of the candle is M
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v =
for when n balls have collided
v = v₀
Now let's analyze the case of the bouncing ball (elastic)
initial instant
p₀ = m v₀
final moment
p_f = m v_{1f} + M v_{2f}
p₀ = p_f
m v₀ = m v_{1f} + M v_{2f}
m (v₀ - v_{1f}) = M v_{2f}
this case corresponds to an elastic collision whereby the kinetic energy is conserved
K₀ = K_f
½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²
v₁ = v_{1f} v₂ = v_{2f}
m (v₀² - v₁²) = M v₂²
let's use the identity
(a² - b²) = (a + b) (a-b)
we write our equations
m (v₀ - v₁) = M v₂ (1)
m (v₀ - v₁) (v₀ + v₁) = M v₂²
let's divide these equations
v₀ + v₁ = v₂
Let's look for the final speeds
we substitute in equation 1
m (v₀ - v₁) = M (v₀ + v₁)
v₀ (m -M) = (m + M) v₁
v₁ = v₀
we substitute in equation 1 to find v₂
v₂ = v₀ -
v₀
v₂ =
v₂ =
v₂ =
Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.
In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.
Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.