Wednesday

July 30, 2014

July 30, 2014

Posted by **Elizabeth** on Monday, October 22, 2012 at 8:22pm.

Answer will be in kg.

- Physics -
**Elena**, Tuesday, October 23, 2012 at 10:31amChange 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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