phy mock question
- 天同Lv 78 年前最愛解答
3. Yes, you are right. The kinetic energy on jumping upward needs to be included.
4. (a) Observer on ground: the ball moves forward and falls in a parabolic trajectory.
Observer on toy are: the ball falls backward following a (nearly) parabolic trajectory.
(b)(i) acceleration of toy car = 10/4 m/s^2 = 2.5 m/s^2
Use v^2 = u^2 + 2a.s to find speed of toy car after travelling 2 m
v^2 = 0^2 + 2 x 2.5 x 2 (m/s)^2
v^2 = 10 (m/s)^2
v = 3.162 m/s
Consider the vertical motion of the ball after it is released,
use: s = ut + (1/2)at^2
with s = 0.5 m, u = 0 m/s, a = -g(=-10 m/s^2), t=?
0.5 = (1/2).(10)t^2
t = 0.3162 s
It takes 0.3162 s for the ball to land on the toy car
Use: s = ut + (1/2)at^2 to find the distance travelled by the toy car in 0.3162 s
hence, s = [3.162 x 0.3162 + (1/2) x 2.5 x 0.3162^2] m = 1.125 m
(ii) Horizontal distance travelled by the falling ball
= 3.162 x 0.3162 m = 1 m
Hence, s = (1.125 - 1) m = 0.125 m