Sally and Suzy are moving into their first college dorm together. They are loading all their furniture onto a truck with a ramp
1 answer:
You might be interested in
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.
PART a)
here when stone is dropped there is only gravitational force on it
so its acceleration is only due to gravity
so we will have
Part b)
Now from kinematics equation we will have
now we have
y = 25 m
so from above equation
Part c)
If we throw the rock horizontally by speed 20 m/s
then in this case there is no change in the vertical velocity
so it will take same time to reach the water surface as it took initially
So t = 2.26 s
Part D)
Initial speed = 20 m/s
angle of projection = 65 degree
now we have
PART E)
when stone will reach to maximum height then we know that its final speed in y direction becomes zero
so here we can use kinematics in Y direction
so it will take 1.85 s to reach the top