A block of mass m slides on a frictionless horizontal surface with initial speed v₀ at t = 0. It experiences a resistive force from the surrounding air of magnitude bv, directed opposite to its velocity, where b is a positive constant.

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Key answer: 1 point: Newton's second law gives m dv/dt = −bv 1 point: separates variables, dv/v = −(b/m)dt 1 point: integrates from v₀ at t = 0 to v at t, giving ln(v/v₀) = −bt/m 1 point: Newton's second law gives m dv/dt = −bv 1 point: separates variables, dv/v = −(b/m)dt 1 point: integrates from v₀ at t = 0 to v at t, giving ln(v/v₀) = −bt/m 1 point: v(t) = v₀e^(−bt/m) 1 point: x = ∫₀^∞ v₀e^(−bt/m) dt 1 point: evaluates to v₀(m/b)[−e^(−bt/m)]₀^∞ 1 point: x = mv₀/b, a finite distance 1 point: speed starts at v₀, decreases, is concave up, and approaches zero asymptotically without reaching it 1 point: position starts at zero with initial slope v₀ and increases 1 point: position approaches the horizontal asymptote x = mv₀/b 1 point: as the block slows, the resistive force also falls in proportion, so the deceleration decreases and the block never quite stops within the model 1 point: the distance covered in each successive interval of time falls off exponentially 1 point: the sum of those ever-smaller contributions converges, so the total distance is finite even though the time taken is unbounded — a physical situation in which "never stops" and "travels a finite distance" are both true

AP Physics C: Mechanics · Question 10

A block of mass m slides on a frictionless horizontal surface with initial speed v₀ at t = 0. It experiences a resistive force from the surrounding air of magnitude bv, directed opposite to its velocity, where b is a positive constant.

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