a) h = v0^2 sin^2(θ) / 2g
b) (M + m) v = M v0 cos(θ) + m v
c) d = (v^2 - v0^2 sin^2(θ)) / 2g
d) L = v * sin(θ) * t
(a) Maximum height reached by the skeet before the collision
The maximum height reached by the skeet before the collision is given by:
h = v0^2 sin^2(θ) / 2g
where:
v0 is the initial speed of the skeet
θ is the angle at which the skeet is fired
g is the acceleration due to gravity
(b) Velocity of the skeet-pellet system right after the collision
The magnitude of the velocity of the skeet-pellet system right after the collision can be found using the conservation of momentum:
(M + m) v = M v0 cos(θ) + m v
where:
M is the mass of the skeet
m is the mass of the pellet
v is the velocity of the skeet-pellet system right after the collision
v0 is the initial speed of the skeet
θ is the angle at which the skeet is fired
The direction of the velocity of the skeet-pellet system right after the collision can be found using the conservation of energy:
(M + m) v^2 / 2 = M v0^2 cos^2(θ) / 2 + m v^2 / 2
Solving for v, we get:
v = sqrt(M v0^2 cos^2(θ) + m v^2) / (M + m)
(c) Distance the skeet-pellet system goes higher than h due to the collision
The distance the skeet-pellet system goes higher than h due to the collision is given by:
d = (v^2 - v0^2 sin^2(θ)) / 2g
(d) Horizontal distance traveled by the skeet-pellet system after the collision
The horizontal distance traveled by the skeet-pellet system after the collision is given by:
L = v * sin(θ) * t
where t is the time it takes for the skeet-pellet system to hit the ground. This can be found using the following equation:
t = sqrt(2h / g)
Substituting this into the previous equation, we get:
L = v * sin(θ) * sqrt(2h / g)
Question
A skeet (clay target) of mass M is fired at an angle from the horizontal with an initial speed v0. When it reaches the maximum height h, it is hit from below by a pellet of mass m traveling vertically upward at a speed v (Fig. 1). Right after the collision, the pellet is embedded in the skeet.
a- Determine the maximum height h reached by the skeet before the collision.
b- Determine the magnitude and direction of the velocity of the skeet-pellet system right after the collision.
c- By what distance d does the skeet-pellet system go higher than h due to the collision?
d- What is the horizontal distance L traveled by the skeet-pellet system after the collision?