A block of mass m=2.2kg moving on a frictionless surface with a speed Vi= 9.1m/s

Question

A block of mass m=2.2kg moving on a frictionless surface with a speed Vi= 9.1m/s, makes a perfectly elastic collison with a block of mass M at rest.  After the collision, the 2.2 kg block recoils with a speed of Vf=0.7m/s.  In the figure, the mass M is closest to?

The answer is 2.6kg.  

I'm pretty sure I know which equation to use but not how to isolate M.  

Vf=[(m-M)/(m+M)]vi

A block of mass m=2.2kg moving on a frictionless surface with a speed Vi= 9.1m/s, makes a perfectly elastic collison with a block of mass M at rest. After the collision, the 2.2 kg block recoils with a speed of Vf=0.7m/s. In the figure, the mass M is closest to? The answer is 2.6kg. I'm pretty sure I know which equation to use but not how to isolate M. Vf=[(m-M)/(m+M)]vi

Explanation / Answer

By the law of momentum conservation:-
=>m1u1+m2u2 = m1v1+m2v2
=>2.2 x 9.1 + 0 = 2.2 x (-0.7) + Mv2
=>Mv2 = 21.56 N-s ----------(i)
As the collision is elastic:-
=>v1-v2 = u2 - u1
=>-0.7 -v2 = 0 - 9.1
=>v2 = 8.4 m/s ---------(ii)
By putting the value of v2 from (ii) in to (i):-
=>M = 21.56/8.4
=>M = 2.57
=>M ? 2.6 kg

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