In the figure, block 2 of mass 2.10 kg oscillates on the end of a spring in SHM
ID: 1388774 • Letter: I
Question
In the figure, block 2 of mass 2.10 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (0.900 cm) cos(omega t + pi/2). Block 1 of mass 4.20 kg slides toward block 2 with a velocity of magnitude 9.00 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.50 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 0.0237 Units mExplanation / Answer
There is no friction, so that oen can use the conservation energy.
Data are as follows:
m1= 4.20 kg
v1 = 9 m s^-1
m2= 2.10 kg
A2 = 0.9 cm = 9 10^-3 m
P2= 26 ms = 2.6 10^-2 s => omega = 2pi/P2 = 241 rad s^-1. It follows, from k = omega^2 m, that the elastic constant is k = 1.2 10^5 N m^-1.
t_0 = 6.50 ms = 6.5 10^-3 s
The speed of m2 is v2(t) = dx2/dt = A2 omega sin (omega t + pi/2) = 2.16 sin (omega t + pi/2)
At the instant t_0 of the collision, the kinetic energy of m1 is K1 = (1/2) m1 v1^2 = 170.1 J
and the phase of m2 is omega t_0 + pi/2 = 179.7 deg.
v2(0)=4.3 10^-3 m s^-1
x2(0) = -8.99 10^-3 m
The total energy of m2 is E2 = (1/2) m2 v2^2 + (1/2) k x2^2 = 4.8 J (it is mostly potential energy)
After the collision, the kinetic energy of m1 is transferred entirely to m2, which, thus, has a total energy E = K1+ E2 = 175 J. It is E = (1/2) k A^2 => A = sqrt(2 E/ k) = 5.4 10^-2 m = 5.4 cm.