Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
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Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
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