PHY problem plz help for CE

[匿名]

2009-09-11 11:11 am

Position and movement

Q1. a bullet is fired towards a nearby tree with a speed of 200ms^-1
The bullet is later found at depth of 5cm. Find the average deceleration
of the bullet inside the tree?

Q2 A ball is released from rest and falls vertically under gravity. If the distance travelled by the ball in the 1st second is p, and that travelled in the 2nd second is q, what is the ratio of q:p?

At last i want to ask what is meaning of the car during 4th second and also the meaning of 1st second and 2nd second in Q2?Is it suppose to mean t=4s, t=1s and t=2s respectively ?
For all above questions plz give detail explanation THX

回答 (1)

Prof. Physics

2009-09-11 4:32 pm

✔ 最佳答案

1. By equation of motion,

v2 = u2 + 2as

0 = (200)2 + 2a(0.05)

Acceleration, a = -400 000 ms-2

So, the average deceleration is 400 000 ms-2

2. By p = 1/2 gt2

p = 1/2 g(1)2 = g/2

Consider the 2nd second, let the distance travelled after the 2nd second be D

D = 1/2 gt2

D = 1/2 g(2)2 = 2g

So, q = D - p = 3g/2

q : p = 3 : 1

In the 1st second and the 2nd second, it just refers to the motion of the ball within this particular second.

參考: Physics king

👍 0 👎 0