Question

A 0.250 kg skeet (clay target) is fired at an angle of ? = 32.2° to the horizon with a speed of vo = 32.3 m/s, as seen in figure below. When it reaches the maximum height, it is hit from below by a 15 g pellet traveling vertically upward at a speed of v = 194 m/s. The pellet is embedded in the skeet. How much higher did the skeet go up? How much extra distance, ?x, does the skeet travel because of the collision?

Answer 1

First Part -

Consider the ideal situation, means there is no air drag.

Now, at the highest point, the y velocity = 0m/s.
Using momentum,
(0.25)(0) + (15/1000)(194) = (0.015+0.25)(v)

=> v = (0.015 x 194) / 0.265 = 10.98 m/s
use the expression -

v^2 = u^2 + 2as
0 = 10.98^2 + 2(-9.8)(s)

=> s = 10.98^2 / (2 x 9.8) = 6.15 m higher up.

Second Part -

X- component of velocity = 32.3 cos 32.2 = 27.3 m/s

Y - component of velocity = 32.3 sin32.2 = 17.2 m/s

Now, time taken by the skeet to cover extra height s = 6.15 m

s = ut + (1/2)*a*t^2

=> 6.15 = 10.98t - 0.5*9.8*t^2

=> 4.9t^2 - 10.98t + 6.15 = 0

=> t = [10.98 + sqrt(10.98^2 - 2*4.9*6.15)] / (2*4.9) = [10.98 + 0.14] / 9.8 =1.13 s

other value of t = [10.98-0.14] / 9.8 = 1.11 s

So, the extra distance travelleed by the skeet, delta x = 1.13 x 27.3 = 30.85 m

and the other value of delta x = 1.11 x 27.3 = 30.3 m

so, our answers are 30.85m or 30.3 m.