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Correct Answer: 17.14 ms-1
to solve the problem of determining the speed of the remaining parts of a rock after an explosion, we can use the principle of conservation of momentum. according to this principle, the total momentum of a closed system remains constant if no external forces are acting on it.
initially, the rock is at rest, and hence its total initial momentum is zero. let's denote the mass of the rock as \( m = 400 \) kg. after the explosion, 30% of the rock's mass is thrown away at a speed of 40 m/s. let's calculate the mass of the part thrown away and the remaining mass:
- mass thrown away \( = 0.3 \times 400 \) kg \( = 120 \) kg.
- remaining mass \( = 0.7 \times 400 \) kg \( = 280 \) kg.
let \( v \) be the velocity of the remaining part of the rock. since the total initial momentum of the system was zero, the total final momentum must also be zero for momentum to be conserved. therefore, we set up the momentum conservation equation:
\[ 0 = (mass \, of \, thrown \, part \times speed \, of \, thrown \, part) + (mass \, of \, remaining \, part \times speed \, of \, remaining \, part) \]
\[ 0 = (120 \, kg \times 40 \, m/s) + (280 \, kg \times v) \]
solving for \( v \):
\[ 0 = 4800 + 280v \]
\[ 280v = -4800 \]
\[ v = -4800 / 280 \]
\[ v = -17.14 \, m/s \]
the negative sign indicates that the direction of the velocity of the remaining part is opposite to the direction in which the 30% of the mass was thrown. since the problem only asks for the speed, we consider the magnitude:
\[ |v| = 17.14 \, m/s \]
therefore, the speed of the remaining part of the rock is 17.14 m/s.
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