AP Physics C Unit 4: Linear Momentum

Mass 0.5 kg with velocity components 3.0 m/s and -4.0 m/s has momentum magnitude 2.5 kg*m/s.

At fixed kinetic energy K_0, doubling the mass scales momentum by sqrt(2), which equals 2 p_0 / sqrt(2).

Quadrupling kinetic energy K_0 -> 4 K_0 increases momentum by a factor of sqrt(4) = 2.

Differentiating x(t) = 0.5 t^3 gives v(4.0) = 24 m/s; multiplying by 2.0 kg gives 48 kg*m/s.

Slopes give v_i = +0.5 m/s and v_f = -2.0 m/s; Delta p = 3.0 kg * 2.5 m/s = 7.5 kg*m/s.

Mass doubles while speed is halved, so final momentum remains exactly p_0.

For equal momentum p_0, speed is inversely proportional to mass; since m_1 > m_2, v_2 > v_1.

Velocity is tangent to the semicircular path at crest Y, directed horizontally to the right.

Mass of the cart-water system decreases while speed is constant, so system momentum decreases.

Both receive 10 N*s impulse; the 40-kg student-chair has less mass than 50 kg, so it reaches greater speed.

Rebounding reverses velocity direction, so impulse magnitude is m_1 (v_1f + v_1i).

Padding cushions impact: collision duration increases, maximum force decreases, and area under F_h(t) remains the same.

Integrating F_0 e^{-bt} dt from 0 to infinity gives total momentum F_0 / b.

Trial 2 has greater force at all times, giving greater impulse (J_1 < J_2) and greater final kinetic energy (W_1 < W_2).

The area under a net force-time curve gives impulse, which equals the change in momentum regardless of initial velocity.

Net force is 2kt - mg sin(theta); integrating from t_1 to t_2 gives k(t_2^2 - t_1^2) - mg sin(theta)(t_2 - t_1).

The negative triangle gives J_1 = -0.5 N*s and the positive triangle gives J_2 = +0.5 N*s, so J_1 = -J_2.

Constant mass flow gives constant thrust; with decreasing system mass, acceleration must increase.

More mass is moving across the ice in Scenario 3 (3m vs m in Scenario 1), giving the greatest center of mass speed.

Center of mass speed is unchanged (v_1 = v_2 = 0.5 v_0), and stopping Block A requires more impulse than sticking (J_1 > J_2).

Total initial momentum is m v_0 and total mass is 3m; with zero external force, v_cm = v_0 / 3 at all times.

In all collision outcomes, external force is zero, so center of mass speed is identically conserved: v_Pcm = v_Qcm = v_Rcm.

Slope is m_1 / (m_1 + m_2); replacing m_1 with 2m_1 increases this fraction, making the slope greater.

4 kg moving left vs 3 kg moving right; total momentum sign depends on whether v_3 / v_4 is greater or less than 4/3.

Total momentum is conserved at 18 kg*m/s and relative speed is 2 m/s before and after, matching Choice A.

Conserving momentum in x and y gives v_x = v_0(1 - cos theta) and v_y = -v_0 sin theta.

Recoil speed of 3m block is v/3; relative speed is 4/3 v, giving separation D + 4/3 v (t - t_1).

Elastic collision conserves kinetic energy: 72 J = 2 v^2 + 6 J yields v = sqrt(33) approx 5.7 m/s.

At turning point Q the box is at rest (v = 0), so zero momentum and zero kinetic energy are identically preserved.

Combined mass 0.75 kg moves at 4/3 m/s; K_f = 0.5 * 0.75 * (4/3)^2 = 2/3 J.

In elastic Collision 1, Cart Y gets all K_i; in inelastic Collision 2, Cart Y gets less than K_i, so speed is greater after Collision 1.

Kinetic energy lost in a sticky collision is 0.5 * mu * v_rel^2; since mu and v_rel are identical, the ratio is 1.

Momentum gives v_2 = v/2; computing K_f shows K_f = K_i, so the collision is confirmed elastic, matching Choice B.

K_i = 162 J and K_f = 54 J; the decrease in kinetic energy is 162 J - 54 J = 108 J.

Graph B shows equal masses completely exchanging momentum of 4.8 kg*m/s, conserving total kinetic energy.

v_f = (m_1 v_1) / (m_1 + m_2) is an inverse function: decreasing, concave up, asymptotically approaching zero (Graph A).