Paper 2 Dynamics Answers

Q1 Connected bodies

1. Combined mass $=5.0\mathrm{kg}$. $a=F/m=15.0/5.0=3.0\mathrm{m}{\mathrm{s}}^{-2}$. [2]
2. For the $2.0\mathrm{kg}$ block, the only horizontal force is tension:

$$T=ma=2.0(3.0)=6.0\mathrm{N}.$$

[3] 3. The tension forces are internal to the two-block system. They are equal and opposite forces between parts of the chosen system, so they cancel in the system equation. [2]

Q2 Lift motion

1. $N=mg=58.0(9.81)=569\mathrm{N}$. [1]
2. Upward positive: $N-mg=ma$, so

$$N=m(g+a)=58.0(9.81+1.20)=639\mathrm{N}.$$

[3] 3. For downward acceleration, $N=m(g-a)=58.0(9.81-1.20)=499\mathrm{N}$. [3] 4. In ideal free fall, the student and scale both accelerate downward at $g$. No contact force is needed between them, so the normal force and scale reading are zero, although weight still acts. [2]

Q3 Impulse

1. $p_i=0.060(24)=1.44\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$. $p_f=0.060(-18)=-1.08\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$. Hence $\Delta p=-1.08-1.44=-2.52\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$. [3]
2. $\Delta t=4.0\times 10^{-3}\mathrm{s}$:

$$F_{\mathrm{avg}}=\frac{\Delta p}{\Delta t}=\frac{-2.52}{4.0\times 10^{-3}}=-630\mathrm{N}.$$

Magnitude $=630\mathrm{N}$. [2] 3. Opposite to the initial direction of the ball. [1] 4. They are a Newton’s third-law pair: the two forces are from the same interaction, act on different bodies and are equal in magnitude and opposite in direction. [2]