In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in
ID: 2069076 • Letter: I
Question
In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos(?t + ?/2). Block 1 of mass 3.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?
In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos(?t + ?/2). Block 1 of mass 3.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?Explanation / Answer
T=20ms
=2/T
=100
v=0.01**sin(t+/2)
vmax= 0.01
energy balancing gives
1/2*k*xmax^2= 1/2*m*vmax^2
k= 4*(0.01)^2/(0.01)^2
k=394784.176
at t=5.0
x=0.01 cos(/2+/2)
x= -0.01m
v=0
intial momentum= 3*6=18
final momentum= (4+3)v= 18
v=18/7 m/sec
now lets say it moves x1 distance more
1/2 k*x1^2 = 1/2(m1+m2)*v^2
394784.176*x1^2= 7*(18/7)^2
x=1.082 cm
so final amplitude= 1+1.083 cm
=2.083 cm