Physics....唔知點計....附有答案

6.a. Net force acting on the smaller car = 4000 – 400 = 3600 N
Work done = Fs = 3600X 50 = 180 000 J
b. The work done by the net force on the smaller car is converted to the kinetic energy of it. So kinetic energy = 180 000 J
c. By K.E. = 1/2 mv2
180 000 = 1/2 (1200)v2
Speed, v = 17 ms-1

21. a. By conservation of momentum
0 = mava + mbvb
0 = (50)(1.2) + mb(-1.5)
Mass of Lauren, mb= 40 kg
b. Impulse = change of momentum
impulse = mava - maua
impulse = (50)(1.2 -0)
impulse = 60 Ns
c. As no external force acts in the system, momentum is conserved. Hence impulse acting on Lauren= 60 Ns
d. Total momentum is conserved. Hence total momentum = initial total momentum = 0.
e. Their final velocities would be larger. It is because they experience a larger impulse, by impulse = Force X time, for the same impact time, the larger the force, the larger the impulse, and the larger the change in momentum and hence velocity.

11.a. Impulse = Force X time = 10 X 6 = 60 Ns
b. Impulse in the latter 6 s = 5 X 6 = 30 Ns
So, total change in momentum = 90 kgms-1
Since at first the object was at rest, hence momentum after 12 s is 90 kgms-1.
c. At the first 6 s,acceleration of the object = F / m = 10 / 2.5 = 4 ms-2
In the latter 6 s,acceleration = F’ / m = 5 / 2.5 = 2.0 ms-1

圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn02Apr191739.jpg?t=1208598253


13.a. Impulse = area under Force-time graph
~ 3200 X 0.10 / 2
= 160 Ns
b. impulse = change of momentum
160 = 60v – 0
v = 2.7 ms-1