Question

A block of mass 1.6 ?? is moving across a smooth floor at 13.8 ?/? and encounters a second block (initially at rest) of mass 3.4 ?? in a fully elastic collision. The second block is attached to a spring of ? = 1250 ?/?. Assume the spring to be massless and does not interfere with the collision. After the collision, the second block is under simple harmonic motion. Determine, a. The amplitude of oscillation. b. The frequency of oscillation. c. The time it takes to go from equilibrium to the maximum displacement. d. Functional form of SHM (i.e. ? ? = ????(?? + ?)

Answer 1

V20 13.8 m/s 1.6kg 3.4kg After collision oV₂ 1.org 1 3.4 kg conservation of momentum . 1-6x 13.8 = 16v, +3.4 V2 By → 1.6V, + 3.4 72 = 22.08 -0 also V₂-No= e=1 13.8 soning, we get, V₂ = 8,83 m/s

= 19.2 rad/s

w = Vk/m= 1250/3.4

A)

V = Au

8.83 = A × 19.2

A = 0.46 m = 46 cm

(B) = 19.2/(2π) =3.06 Hz

f=w/(27)

(C) Time period

= 2π/19.2 = 0.328 s

T = (21)

t = T/4 = 0.082 s

(D) taking initial position as x=0,

r = Asin(wt)

x = (0.46m) sin (19.2t)