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A block of mass = 3.6 kg, moving on a frictionless surface with a speed vi = 9.3.....

A block of mass m=3.6 kg, moving on a frictionless surface with a speed vi = 9.3 m/s, makes an elastic collision with a block or mass M at rest. After the collision,the 3.6 kg block recoils with a speed of v1 = 2.7 m/s. The speed of the other block after the collision is closest to: A) 6.6 m/s B) 8.0 m/s C) 9.3 m/s D) 10.7 m/s E)12.0 m/s
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Answer #1
by conservation of momentum we have
mu=-mv+MV
m=3.6
u=9.3
v=2.7
V=43.2/M
by conservation of energy
we have
.5mu2=0.5mv2+0.5MV2
by solving the eqns we get
V=6.6m/s
answered by: mathhater2192
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