Question

In the figure, block A (mass 1.0 kg) slides into block B (mass 2.3 kg), along a frictionless surface. The directions of velocities before and after the collision are indicated; the corresponding speeds are vAi = 5.9 m/s, vBi = 2.5 m/s, and vBf = 4.4 m/s.?

Answer 1

I found that you have an inelastic collision where mass `A` moves to the right with velocity of `1.5m/s`

Explanation:

Considering:

I would start using conservation of momentum that works for both elastic or inelastic collisions:
`vecp_("before")=vecp_("after")`
`m_Av_(Ai)+m_Bv_(Bi)=m_Av_(Af)+m_Bv_(Bf)`
`1*5.9+2.3*2.5=1*v_(Af)+2.3*4.4`
`v_(Af)=11.65-10.12=1.5m/s` in the same direction as `v_(Bf)` (to the right in figure).

We can check whether is elastic or not observing if Kinetic Energy, `K`, is conserved:
`K_("before")=K_("after")`
`1/2m_Av_(Ai)^2+1/2m_Bv_(Bi)^2=1/2m_Av_(Af)^2+1/2m_Bv_(Bf)^2`
cancel the `1/2` and substitute the values:
`1*5.9^2+2.3*2.5^2=1*1.5^2+2.3*4.4^2`
`49.2=45.8` No...it is not prfectly elastic. Some energy is lost after the impact.