In fig. 15-40, block 2 of mass 2.0 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 (ωt + π/2). block i of mass 4.0 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?
fig. 15-40

Amplitude= -0.03087m

Explanation

Note that the total momentum before collision is equal to the momentum after collision.also

Energy Before Collision = Energy After Collision

the parameters are as follows

m1=6kg

m2=2kg

T=20ms=.02s

v1=6m/s

t=5ms=.005

x = (1.0 cm) cos (ωt + π/2)

There are some other equation tat could be useful

ω=2πf

ω=

f=1/T

kinetic energy is Ek=0.5mv^2

Energy in spring=Es=0.5kx^2

Lets compute energy before collision

kinetic energy of block 1 =0.5m1v1^2

energy of block 2 which is   a spring is =0.5kx^2

to find k, recall that  ω=

We can find ω=2πf ,

f=1/T    ,f=1/0.02=50HZ

2*PI*50

ω=100πrad/s

ω=

ω^2m2=k    ........find the square root of both sides

k=197317

to find x, remember the position of the spring

x = (1.0 cm) cos (ωt + π/2)

ω=100πrad,  t=0.005s

x = (0.01) cos (100πrad*0.005s + π/2)=

-0.01m

energy of block 1=0.5*4*6=72J

ENERGY OF SPRING=0.5*2*(100π)^2 -0.01^2=9.86J

energy before collision will be=e1+e2=  72+9.86

81.86J

energy after collision will be  Eafter=0.5kx^2

0.5kx^2=0.5m*ω^2 *x^2

m=m1+m2= 6kg

ω==(197317/6)^0.5

181.3rad/s

Energy Before Collision = Energy After Collision

81.86J=(0.5)*(6)*181.3rad/s^2(x^2)

x=0.0288m

Amplitude can now be found since we have gotten all the parameters

x=A cos (ωt + π/2)

0.0288m=A cos (181.3rad/s*(0.005)+ π/2)

A= -0.03087m

massxgravity x heght=weightxheight= pe

4900=490xh

h=10

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west