Q3. A skeet (a small disc) of massM is fired at an angle � to the horizontal with a speed v0. When
it reaches the maximum height it is hit from below by a pellet of mass m (m < M) traveling
vertically upward at a speed of v. The pellet is embedded in the skeet.
(a) What is the horizontal velocity of the skeet immediately after collision?
(b) What is the vertical velocity of the skeet immediately after collision?
(c) How much higher did the skeet go up i.e., calculate h′?
(d) How much extra distance Δx does the skeet travel because of the collision?
Q
To find the answers to these questions, we need to analyze the conservation of momentum and energy before and after the collision between the skeet and the pellet.
(a) To find the horizontal velocity of the skeet immediately after the collision, we need to consider the conservation of horizontal momentum. Before the collision, the horizontal velocity of the skeet is given by v0*cos(theta), where theta is the launch angle. Since there are no external horizontal forces acting on the system, the horizontal momentum is conserved after the collision as well. Therefore, the horizontal velocity of the skeet immediately after the collision is also v0*cos(theta).
(b) To find the vertical velocity of the skeet immediately after the collision, we need to consider the conservation of vertical momentum. Before the collision, the vertical velocity of the skeet is given by v0*sin(theta). After the collision, the pellet is embedded in the skeet, adding its mass and velocity in the upward direction. The vertical momentum is conserved before and after the collision, so we can write:
M * v0 * sin(theta) = (M + m) * vf,
where vf is the vertical velocity of the skeet immediately after the collision. Solving for vf:
vf = (M * v0 * sin(theta)) / (M + m).
(c) To find how much higher the skeet goes up after the collision, we need to calculate its maximum height h' after the collision and subtract the initial maximum height h. The initial maximum height h can be determined using the kinematic equation:
h = (v0^2 * sin^2(theta)) / (2 * g),
where g is the acceleration due to gravity. After the collision, the vertical velocity of the skeet becomes vf, and using another kinematic equation, we can find the maximum height h':
vf^2 = 0 - 2 * g * h',
Solving for h':
h' = vf^2 / (2 * g).
(d) To find the extra distance the skeet traveled because of the collision, we need to calculate the horizontal displacement covered during the time the skeet is at maximum height before the collision and after the collision. The horizontal displacement is given by:
delta_x = v0 * cos(theta) * t,
where t is the time taken to reach maximum height. Using the equation of motion for vertical motion:
vf = 0 - g * t,
Solving for t:
t = vf / g.
Substituting the value of t in the horizontal displacement equation:
delta_x = v0 * cos(theta) * (vf / g).
To find the extra distance traveled, we subtract the displacement before the collision from the displacement after the collision:
Delta_x = delta_x - delta_x_before_collision.
Note that delta_x_before_collision is the horizontal displacement covered during the time the skeet is at maximum height before the collision, which can be calculated using the initial vertical velocity and the time of flight t:
delta_x_before_collision = v0 * cos(theta) * (v0 * sin(theta) / g).
So, the extra distance traveled is given by:
Delta_x = v0 * cos(theta) * (vf / g) - v0 * cos(theta) * (v0 * sin(theta) / g).
Simplifying the equation gives:
Delta_x = v0 * cos(theta) * (vf / g - v0 * sin(theta) / g).
To answer these questions, we'll start by analyzing the conservation of momentum and conservation of energy during the collision.
(a) First, let's calculate the horizontal velocity of the skeet immediately after the collision.
Since the pellet is traveling vertically upward, its horizontal velocity is zero (v_x = 0).
The horizontal velocity of the skeet before the collision is given by v_x0 = v_0 * cos(θ).
Since momentum is conserved in the horizontal direction, the horizontal velocity of the skeet after the collision will also be v_x0.
(b) Now let's calculate the vertical velocity of the skeet immediately after the collision.
The vertical velocity of the pellet before the collision is v_y = v.
The vertical velocity of the skeet before the collision is given by v_y0 = v_0 * sin(θ).
Since momentum is conserved in the vertical direction, m * v_y + M * v_y0 = (m + M) * v_y',
where v_y' is the vertical velocity of the skeet after the collision.
Now, substitute the values: m * v + M * v_0 * sin(θ) = (m + M) * v_y'.
Rearranging the equation, we get v_y' = (m * v + M * v_0 * sin(θ)) / (m + M).
(c) To find out how much higher the skeet goes up, we need to calculate the difference in the maximum heights before and after the collision.
Before the collision, the skeet reaches a maximum height h_0.
After the collision, when the skeet reaches the maximum height, its vertical velocity is momentarily zero.
Using the kinematic equation v_y^2 = v_y0^2 - 2 * g * h, where g is the acceleration due to gravity,
we can find the height h' using the vertical velocity v_y' calculated in part (b).
v_y'^2 = v_y0^2 - 2 * g * h'.
Substituting the values, (m * v + M * v_0 * sin(θ))^2 / (m + M)^2 = (v_0 * sin(θ))^2 - 2 * g * h'.
Rearranging the equation, we get h' = [(m * v + M * v_0 * sin(θ))^2 / (m + M)^2 - (v_0 * sin(θ))^2] / (2 * g).
(d) To find the extra distance traveled by the skeet due to the collision, we need to calculate the horizontal displacement.
The horizontal displacement is given by Δx = v_x * t, where t is the time taken for the skeet to reach the maximum height after the collision.
Since the vertical velocity becomes momentarily zero at the maximum height after the collision, the time taken can be calculated using the equation v_y' = v_y0 - g * t.
Substituting the values, (m * v + M * v_0 * sin(θ)) / (m + M) = v_0 * sin(θ) - g * t.
Rearranging the equation, we get t = [(m * v + M * v_0 * sin(θ)) / (m + M)] / g.
Now, substitute this value of t in the equation Δx = v_x * t to calculate Δx.
Note: In these calculations, θ represents the angle of projection.