A 0.40-kg skeet (clay target) is fired at an angle of 28 to the horizon with a s

Question

A 0.40-kg skeet (clay target) is fired at an angle of 28 to the horizon with a speed of 25 m/s. When it reaches the maximum height, h, it is hit from below by a 26-g pellet traveling vertically upward at a speed of230m/s. The pellet is embedded in the skeet.

Part A

How much higher, h?, did the skeet go up?

Express your answer to twoe significant figures and include the appropriate units.

Part B

How much extra distance, ?x, does the skeet travel because of the collision?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Maximum height reached by a skeet=h= v^2.sin^2tetha/2g

= (25)^ 2 (sin^2 28 /2×9.8) = 7.028m

momentum of pellet = momentum of (pellet +skeet)

0.026×230=(0.026+0.40)v

5.98=0.426v

v=14.03m/s

Actual range of projectile=R=v^2.sin^2tetha/g

= (25)^2 sin 2*28 /9.8 = 52.87m

H=1/2gt^2

0.5 x 9.8 t^2 = 8.65+7.028

t = 1.79 s

velocity of the skeet along x direction = 25 cos 28

Initial momentum of the skeet along x direction = final momentum of the skeet + pellet

0.25 (25 cos28) = (0.25+0.015)Vx

Vx= 20.8 m/s

X = Vx t = (20.8)* 1.79 = 37 m

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