A0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. when it reaches the maximum height, it is hit from below by a 15 g pellet travelling upward at a speed of 200 m/s. the pellet is embedded in the skeet. (a) how much higher did the skeet go up? (b) how much extra distance does the skeet travel because of the collision?

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                          

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

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