Homework Answers
speed of m1 when it hits m2
v = sqrt(2gh) = sqrt(2*9.8*5)
= 9.9 m/s
let the speed after collision be v1 and v2
by the conservation of momentum we have
4.92*9.9 + 0 = 4.92*v1 + 9.4v2 -------(1)
velocity of approach = velocity of recess
v2 - v1 = u1 - u2
v2 - v1 = 9.9 ---------(2)
multiply (2) by 4.92 and adding we have
14.32 v2 = 97.416
v2 = 6.8 m/s
v1 = 6.8-9.9 = -3.097 m
so the height reached by m1 after collision = v1^2 /2g
= (3.097^2) / 2*9.8 = 0.49 m
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