Question #84d11

1 Answer
May 12, 2016

#0.05m#

Explanation:

First of all lets us find out the velocities of both the blocks after first block hits the second and also compresses the spring.

Initial momentum #=m_A xx v_(Ai)#, second block being at rest

Initial momentum #=2 xx 0.15=0.3kgms^-1#
Assuming that the blocks keep on moving in the same direction due to coupling provided by the spring

Final momentum #=m_A xx v_(Af)+m_B xx v_(Bf)#
Equating the two due the conservation of momentum we get
#m_A xx v_(Af)+m_B xx v_(Bf)=0.3#, inserting values of masses

#2 xx v_(Af)+3 xx v_(Bf)=0.3# ......(1)
In the state of equilibrium both the blocks must move at the same speed as an integrated system on a friction-less horizontal surface.
#=>v_(Af)= v_(Bf)#. Let this be equal to #v_f#
Now (1) reduces to
#(2 +3) v_f=0.3#
or #(2 +3) v_f=0.3#
#v_f=0.3/5=0.06ms^-1#
Using law of conservation of energy

Initial KE #=# Final KE #+# PE of the spring
#1/2m_Av_(Ai)^2=1/2m_Av_(f)^2+1/2m_Bv_(f)^2+1/2kx^2#,
where #k# is the spring constant and #x# is its compression.
Multiplying both sides by #2# and inserting given values we get

#2v_(Ai)^2=(2+3)v_(f)^2+10.8x^2#
#2xx0.15^2=5xx0.06^2+10.8x^2#
or #10.8x^2=0.045-0.018#
or #10.8x^2=0.027#

Solving for #x#
#x=sqrt(0.027/10.8)#
#x=0.05m#