Answer:
6.51 m and 37.13 m
Explanation:
from the question we were given
mass of skeet = 0.25 kg
speed of skeet = 25 m/s
angle = 30 degrees to the horizon
mass of pellet = 15 g
speed of pellet = 2000 m/s
without being hit by the pellet, the x and y components of the skeet velocity are
Vx = 25 cos 30 = 21.65
Vy = 25 sin 30 = 12.5
now
V = U + (a x t)
where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance
-25 sin 30 = 25 sin 30 + (-9.8 x t)
-12.5 = 12.5 - 9.8 t
t = 2.55 s
also
V^2 = U^2 + 2as ( s = vertical distance and V = 0 )
0 = (25 sin 30)^2 + 2 x (-9.8) x Y
19.6 Y = 156.25
Y =7.97 m
the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time
distance = 21.65 x 2.55 = 55.2 m
applying the conservation of linear momentum
on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx) ...equation 1
on the y axis : (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2
(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx
Vx = 20.42m/s
0 + (0.015 x 200) = (0.25 + 0.015) Vy
Vy = 11.32 m/s
now V^2 = U^2 + 2 as
0 = 11.3^2 + (2 x (-9.8) x s)
s = 6.51 m
- to find the extra distance moved after collision we apply
s = ut + 1/2 at^2
-7.98 = 11.32t + 1/2 (-9.8) t^2
4.9 t^2 - 11.32t + 7.98
t = 3.17 s
recall that distance = speed x time
distance = 20.42 x 3.17 = 64.73 m
the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2
= 27.6 m
therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m
is the potential difference applied. So, U depends on