Block 2 of mass 1.90kg oscillates on the end of a spring in SHM with a period of
ID: 1559846 • Letter: B
Question
Block 2 of mass 1.90kg oscillates on the end of a spring in SHM with a period of 12.00 ms. The position of the block is given by x = (1.60 cm) cos(omega t + pi/2). Block 1 of mass 3.80 kg slides toward block 2 with a velocity of magnitude 3.60 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 3.00 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? Number 0.0305 Units m the tolerance is +/-2%Explanation / Answer
from the given data
angular frequency,
w = 2*pi/T = 2*pi/(12*10^-3) = 523.6 rad/s
let k is the spring constant.
w = sqrt(k/m2)
==> k = w^2*m2
= 523.6^2*1.9
= 520898 N/m
we know, v = dx/dt
= 1.6*(-sin(w*t + pi/2))*w
= -1.6*523.6*10^-2*sin(523.6*t + pi/2)
= -8.38*sin(523.6*t + pi/2)
at t = 3 ms
v = -8.38*sin(523.6*3*10^-3 + pi/2)
= 0
Apply conservation of momentum
let v is the speed of both objects after the collision,
1.9*3.6 = (1.9 + 3.8)*v
==> v = 1.9*3.6/(1.9 + 3.8)
= 1.2 m/s
now use, Vmax = A*w
Vmax = A*sqrt(k/(m1+m2))
==> A = Vmax/sqrt(k/(m1+m2))
= 1.2/sqrt(520898/(3.8+1.9))
= 0.00397 m