Question

Question #84d11

Answer 1

`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`