Physics---15 point

John

2010-09-26 9:04 am

An arrow is fired with a speed of 20.0 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 1550 m/s^2 and the block's acceleration has a magnitude of 450m/s^2.

(a) How long does it take for the arrow to stop moving with respect to the block?

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

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

Please show all the stepes clearly, thank you.

回答 (1)

天同

2010-09-26 9:21 pm

✔ 最佳答案

(a) Consider first the block. Apply equation of motion: v = u + a.twith u = 0 m/s, a = 450 m/s2,
hence, v = 450t

Consider the arrow. Apply equation of motion: v = u + at
with u = 20 m/s, a = -1550 m/s2hence, v = 20 - 1550tSince both the block and arrow eventually have the same common velocity v, thus,450t = 20 - 1550tsolve for t gives t = 0.01 s

(b) The common speed v = 450 x 0.01 m/s = 4.5 m/s

(c) Aplly equation of motion: s = (1/2).(u + v).t to the arrow
with u = 20 m/s, v = 4.5 m/s, t = 0.01 s, s = s1 (the distance moved by the arrow relative to the surface after striking the block)
hence, s1 = (1/2).(20 + 4.5).(0.01) m = 0.1225 m

Aplly equation of motion: s = (1/2).(u + v).t to the block
with u = 0 m/s, v = 4.5 m/s, t = 0.01 s, s = s2 (the distance moved by the block on the surface after striking by the arrow).
hence, s2 = (1/2).(4.5).(0.01) m = 0.0225 m

Therefore, distance penetrated = (0.1225 - 0.0225) m = 0.1 m

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