Physics

2

(a) Let the direction to the wall is positive
The impulse exerted on the wall = Ft
Ft = mv-mu
= 0.5[(-10)-10]
= -10Ns

(b) The average force exerted on the wall by the ball = F
F = (mv-mu)/t
= -10/0.0025
= -4000N

(c) The impulse received by the player = Ft
Ft = mv-mu
= 0.5(-10-0)
= -5Ns

(d) Apply the equation s = (u+v)t/2
-0.5 = (-10+0)t/2
1 = 10t
t = 0.1s
The average force exerted on the player by the ball = F
F = (mv-mu)/t
= 0.5[0-(-10)]/0.1
= 50N

2007-05-22 00:23:18 補充：
But I cannot understand the direction of the ball.

a)0.5x10-0.5x10
=0kgms-1
b)F=mv-mu/t
=0.5x10-0.5x10/0.0025
= 0 N
c)impulse=mv-mu(0.5x10-0.5x10)
=0.5x0-0.5x10
=-5kgms-1
d)s=1/2(u+v)t
=0.1s
F=mv-mu/t
= 0-5/0.1
=-50N

(a)impulse
=mv-mu
=(0.5kg)(10m/s)-(0.5kg)(-10m/s)
=10kgm/s

(b)Fnet
=(impulse) / (time of impact)
=(10kgm/s) / (0.0025s)
=4000N

(c)impulse
=mv-mu
=(0.5kg)(0m/s)-(0.5kg)(10m/s)
= -5kgm/s(negative sign means the impulse has the opposite direction as the net force)

(d) u=10m/s
v=0m/s
s=0.5m

applied v*v-u*u=2as,
a= -100m/(s*s) (negative sign means deceleration)

applied s=ut+(1/2)at*t,
t=0.1s(repeated)

Fnet
=(impulse) / (time of impact)
=(-5kgm/s) / (0.1s)
=-0.5N
Therefore the required force exerted on the player by the ball is 0.5N.

2007-05-28 14:19:22 補充：
Force 要指明係Fnet 架......