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A 85-kg physics student jumps from a dock into a 135 kg boat that is at rest by the dock. If the velocity

of the student is 4.3 m/s as she leaves the dock, what is the final velocity of the student and the boat?

1 Answer

5 votes

Answer:

The final velocity of the student and the boat is 1.66 m/s

Step-by-step explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is P=mv. If we have a system of bodies, then the total momentum is the sum of them all


P=m_1v_1+m_2v_2+...+m_nv_n

Let's call
m_s = 85\ kg the mass of the physics student,
m_b = 135\ kg the mass of the boat,
v_(so) =4.3\ m/s the initial speed of the student,
v_(bo) =0 the initial speed of the boat,
v_f the final speed of both, assumed common since they keep joined after the jump. Applying the law of conservation of momentum, being
P_o\ and\ P_f the initial and final momentum of the system, we have


P_o=P_f


m_sv_(so)+m_bv_(bo)=(m_s+m_b)v_f

Solving for
v_f


v_f=(m_sv_(so)+m_bv_(bo))/((m_s+m_b))


v_f=(85(4.3)+0)/((85+135))


v_f=1.66 \ m/s

The final velocity of the student and the boat is 1.66 m/s

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