An arrow is fired with a speed of 16.0 m/s at a block of Styrofoam resting on a

Question

An arrow is fired with a speed of 16.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1500 m/s2 and the block's acceleration has a magnitude of 450 m/s2
(a) How long does it take for the arrow to stop moving with respect to the block?
s

(b) What is the common speed of the arrow and block when this happens?
m/s

(c) How far into the block does the arrow penetrate?
m

Explanation / Answer

When the arrow hits the block, it is travelling at 20 m/sec. It then decelerates at 1550 m/s². At the same time, the block itself starts to accelerate at 450 m/s². You want to know when they are travelling at the same speed. So the general equation is: V1 = V0 +a x t Set the two V1s equal to one another, Vo + a x t (arrow) = V0 + a x t (block) 20 m/sec - 1550 m/s² x t = +450 m/s² x t 20 m/sec = 2000 m/s² x t 20 m/sec / 2000 m/s² = t t = 0.01 sec = 10 msec For the common speed, plug in 0.01 sec to the general equation of the arrow (or alternatively, the block): V1 = 20 m/sec - 1550 m/s² x 0.01 V1 = 4.50 m/sec For distance, the equation is d = d0 + V0 x t + 1/2 x a x t² You have to take the distance that the arrow travelled and subtract the distance the block deflected, to find how far the arrow went into the block: d = 0 + (20 x 0.01) + 1/2 x (-1550 x (0.01)²) - (0 + 0 +1/2 x (450 x (0.01)²)) d = 0.2 - 0.0775 - 0.0225 d = .1 m = 10 cm

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