Forces and momentum

1. The student experiences weight mg = 600 N downward and normal force N upward. There is no additional force called acceleration.
2. Taking upward positive, N−600 = 60a gives accelerations +2.0, 0 and −2.0 m s⁻².
3. Weight and normal act on the same body and are not a third-law pair. Each has a partner on a different body.
4. Phase 1 ends at 4 m s⁻¹ and gains ½(2)(2²) = 4 m. Phase 2 gains (4)(3) = 12 m.
5. Phase 3 slows from 4 to 0 m s⁻¹ in 2 s and gains 4 m, so total rise is 20 m.
6. Zero resultant means constant velocity, which may be nonzero. Balanced forces do not require the lift to be stationary.

Answer (a) Accelerations: +2.0, 0, −2.0 m s⁻². The scale-on-student and student-on-scale forces form a third-law pair. (b) Height gained = 20 m; a 600 N reading is consistent with upward constant velocity.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Each nonzero data pair gives D/v² = 0.50 N s² m⁻², supporting quadratic drag over the measured range.
2. Before opening, terminal motion requires 0.50v² = 20, giving √40 m s⁻¹. After opening D = 2.0v², giving √10 m s⁻¹.
3. At 6 m s⁻¹ just after opening, drag is 72 N upward. Net force is 20−72 = −52 N, so a = −52/2 = −26 m s⁻².
4. The package initially still moves downward; the upward acceleration changes its velocity over time rather than resetting it.
5. As it slows, drag falls and the upward resultant becomes smaller, approaching zero at the new terminal speed.
6. The fourfold-drag model assumes the canopy is fully inflated. Inflation time, air density and oscillation can invalidate it during the opening transient.

Answer (a) D = 0.50v²; terminal speeds ≈6.32 and 3.16 m s⁻¹; acceleration = −26 m s⁻² immediately after opening. (b) Finite upward resultant reduces downward speed continuously until drag balances weight.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Initial signed momentum is 2(3)+1(−1) = 5 kg m s⁻¹.
2. External impulse is the triangular signed area, ½(0.20)(−10) = −1.0 N s; final momentum is therefore 4.0 kg m s⁻¹.
3. The stuck mass is 3.0 kg, so v = 4/3 m s⁻¹ east. Internal collision forces cancel in the system sum, but the stated external impulse does not.
4. Initial kinetic energy is ½(2)(3²)+½(1)(1²) = 9.5 J; final is ½(3)(4/3)² = 8/3 J.
5. The reduction is 9.5−8/3 = 41/6 J. The collision may transfer energy to sound, internal energy and deformation.
6. The external force may also do work, which requires displacement information. Impulse alone does not specify work, so the energy loss cannot be attributed entirely to collision deformation.

Answer (a) Final velocity = 4/3 m s⁻¹ east. (b) Kinetic energy falls from 9.5 J to 8/3 J, a reduction of 41/6 ≈6.83 J.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The required momentum-change magnitude is mΔv = 0.20(15) = 3.0 kg m s⁻¹, opposite the initial motion.
2. The triangle area is ½(0.020)(300) = 3.0 N s, consistent with stopping. Average force is 3.0/0.020 = 150 N.
3. For design B, ½(0.060)Fpeak = 3.0, giving 100 N. The average is 50 N, half the peak for a triangular pulse.
4. The ball has the same initial and final momentum, so its impulse cannot change merely by spreading the catch over time.
5. Its kinetic-energy reduction is ½(0.20)(15²) = 22.5 J in both cases. The longer catch can distribute that transfer over a larger distance.
6. Lower peak force can reduce injury or damage, but actual safety also depends on pressure, contact area, material deformation and the available stopping distance.

Answer (a) Impulse magnitude = 3.0 N s; A average = 150 N; B peak = 100 N. (b) Both designs require the same 3.0 N s impulse and remove 22.5 J of ball kinetic energy.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.