Question
how to do 9(b). help
(a) Explain what is the principle of conservation of mechanical energy. (b) An object P of mass 2.0 kg is attached horizontally to one end of a light spring. This spring-mass system is placed on a smooth horizontal plane, as shown in FIGURE 4.7. The spring constant is k = 4.5 kN m^-1. An object Q of mass 3.0 kg moving at a speed of 5.0 m s^-1 collides and sticks to P. It is assumed that there is no dissipation of energy to the surrounding. Find i. the speed of P and Q immediately after the collision. ii. the speed of P and Q at an instant when the spring is compressed by 5.0 cm. iii. the maximum compression experienced by the spring. FIGURE 4.8 shows two objects A and B with each has a mass of 0.5 kg and 1.0 kg respectively. A is released from its rest and hits B causing B to move at a speed of 2.5 m s^-1. Determine the velocity of A immediately after it hits B.
Explanation / Answer
i) Apply law of conservation of momentum
m1u1+m2u2=m1v1+m2v2
3×5=5×v
V=3m/s
ii)Apply law of conservation of energy
V=3m/s let the velocity at that point be v
1/2mv^2+1/2kx^2=1/2×m×V^2
5×3×3-kx^2=5v^2
45-4.5×10^3×0.05×0.05=5v^2
9-2.25=v^2
v=2.6m/s
ìii)1/2×5×v^2=1/2kx^2
5×3×3=4.5×10^3×x^2
x=0.1m
