Posted by Elizabeth on Monday, October 22, 2012 at 8:22pm.
Change of the horizontal component of the stone’s linear momentum is
Δp(x) = m•v(0x) – (-mv(1x)} = m{v(ox)+v(1x)} = m{v(0) •cos15 +v1•cos12}
For the boat Δp=p2-p1 = M•u – 0 = M•u.
Thje law of conservation of linear momentum
Δp(x)= Δp
m{v(0) •cos15 +v1•cos12}= M•u.
M = m{v(0) •cos15 +v1•cos12}/u=
=0.094(14•cos15+10•cos12)/2.2 = 1 kg
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