An ice boat is coasting on a frozen lake. Friction between the ice and the boat

Question

An ice boat is coasting on a frozen lake. Friction between the ice and the boat is negligible, and so is air resistance. Nothing is propelling the boat. From a bridge someone jumps straight down into the boat, which continues to coast straight ahead. (a) Does the total horizontal momentum of the boat plus the jumper change? (b) Does the speed of the boat itself increase, decrease, or remain the same?

(a) yes (b) increase

(a) no (b) remain the same

(a) no (b) increase

(a) yes (b) decrease

(a) no (b) decrease

(a) yes (b) remain the same

(a) yes (b) increase

Explanation / Answer

momentum = mass*velocity in the frictionless situation, the boat's horizontal momentum is conserved since the person who jumps on the boat has no horizontal velocity, the person adds only mass. the momentum before the person jumps on = momentum after the person jumps on, that is (boat mass) * (velocity before) = (boat mass + person mass) * (velocity after) i.e. velocity after = (velocity before) * (boat mass) / (boat mass + person mass) The forward momentum of the boat does not change (by the law of conservation of momentum), but since the mass increases, then to maintain constant momentum, the velocity drops by the ratio of (boat mass) / (boat mass + person mass).

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