Paper 3 Dynamics Answers

Q1 Collision and energy classification

1. Momentum is conserved if the resultant external impulse on the two-trolley system is zero or negligible during the collision. Internal collision forces may be large but cancel within the system. [2]
2. Take the direction of motion as positive:

$$0.80(1.50)+1.20(0.30)=0.80(0.42)+1.20v_B.$$

Initial momentum $=1.56\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$; A final momentum $=0.336\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$. Hence $v_B=1.02\mathrm{m}{\mathrm{s}}^{-1}$. [4] 3. Initial kinetic energy:

$$E_i=\frac{1}{2}(0.80)(1.50{)}^2+\frac{1}{2}(1.20)(0.30{)}^2=0.954\mathrm{J}.$$

Final kinetic energy:

$$E_f=\frac{1}{2}(0.80)(0.42{)}^2+\frac{1}{2}(1.20)(1.02{)}^2=0.695\mathrm{J}.$$

Kinetic energy is not conserved. [4] 4. Inelastic. [1] 5. Momentum conservation depends on external impulse on the chosen system. Kinetic energy can be transformed into internal energy, sound or deformation during the collision even when total momentum is conserved. [2]

Q2 Variable force and impulse

1. Impulse magnitude $=\Delta p=65(5.0)=325\mathrm{N}\mathrm{s}$, opposite the initial motion. [2]
2. Area of triangular force-time graph:

$$\frac{1}{2}(0.200)F_{\max }=325,$$

so $F_{\max }=3250\mathrm{N}$. [3] 3. Impulse equals area under the force-time graph and equals fixed $\Delta p$. Increasing collision time spreads the same impulse over a longer interval, reducing average force and, for similar graph shape, reducing maximum force. [3] 4. Opposite to the runner’s initial velocity. [1] 5. The barrier exerts a force on the runner; the runner exerts an equal and opposite force on the barrier at the same time. These forces act on different bodies and do not cancel on either body’s free-body diagram. [2]