A man with a mass of 55 kg stands up in a \(65-\mathrm{kg}\) canoe of length \(4.0 \mathrm{~m}\) floating on water. He walks from a point \(0.75 \mathrm{~m}\) from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

Initially, the man is 0.75 m from the back of the canoe, and finally, he is 0.75 m from the front of the canoe. So, the man moves a total distance of 4.0 m - 0.75 m - 0.75 m = 2.5 m within the canoe.

Since both the man and the canoe are initially at rest, their initial momentum is 0.

At the end, the man has moved 2.5 m within the canoe. The man and the canoe both have moved, but their total momentum should still be conserved. Let's denote the final velocities of the man and the canoe as v_m and v_c, respectively. The final momentum of the man is the product of his mass and his final velocity: P_m = m_m * v_m, where m_m = 55 kg (mass of the man) The final momentum of the canoe is the product of its mass and its final velocity: P_c = m_c * v_c, where m_c = 65 kg (mass of the canoe)

According to the principle of conservation of momentum, the total momentum before and after the man moves should be equal: Initial momentum = Final momentum => 0 = P_m + P_c => 0 = m_m * v_m + m_c * v_c We need to eliminate one of the unknowns (v_m or v_c) to solve for the other. To do this, we can use the fact that the man moves 2.5 m within the canoe. The man's displacement within the canoe (2.5 m) is equal to the difference between the man's displacement and the canoe's displacement, i.e., 2.5 m = v_m * t - v_c * t, where t is the time it takes for the man to move. Now we have two equations with two unknowns: 1) 0 = m_m * v_m + m_c * v_c 2) 2.5 m = v_m * t - v_c * t Solving these equations together will give us the final velocity of the canoe (v_c).

From equation 2, we can write: v_m * t - v_c * t = 2.5 m => t * (v_m - v_c) = 2.5 m => v_m - v_c = 2.5 m / t Now, replacing v_m in equation 1 with the expression we just found: 0 = m_m * (v_c + 2.5 m / t) + m_c * v_c => 0 = 55 * (v_c + 2.5 m / t) + 65 * v_c To solve for v_c, we can multiply the equation by t, since t cannot be zero: 0 = 55 * t * (v_c + 2.5 m / t) + 65 * t * v_c => 0 = 55 * (t * v_c + 2.5 m) + 65 * t * v_c => 0 = 120 * t * v_c + 55 * 2.5 m Solving for v_c: v_c = (- 55 * 2.5 m) / (120 * t)

Now that we have the final velocity of the canoe (v_c) in terms of the time (t), we can find the displacement of the canoe: Displacement of the canoe = v_c * t => (- 55 * 2.5 m) / (120) = -1.14 m Since the displacement is negative, it means that the canoe has moved 1.14 meters in the opposite direction of the man's movement within the canoe. So, the canoe moves 1.14 meters backward.