Momentum & Collisions

Question 1 (12 marks)

(a) Consider mass $dm = R\,dt$ landing per unit time, brought from $0$ to $v$. Impulse–momentum: $F\,dt = dm\cdot v = (R\,dt)v \Rightarrow F = Rv$. (3, derivation) $F = 4.0 \times 2.5 = 10\ \text{N}$. (2)

(b) Motor power = force × belt speed (force applied moving belt at $v$): $P_{motor} = Fv = 10 \times 2.5 = 25\ \text{W}$. (3)

(c) KE gain rate: each kg gains $\tfrac12 v^2$; rate $= \tfrac12 R v^2 = \tfrac12(4.0)(2.5)^2 = 12.5\ \text{W}$. (2) Difference $= 25 - 12.5 = 12.5\ \text{W}$ dissipated as heat/friction as sand slips and accelerates to belt speed (exactly half the motor power is lost in this "sudden acceleration" process). (2)

Question 2 (12 marks)

(a) Elastic velocity relation: relative speed reverses, so $v_2 - v_1 = u_1 - u_2$. Here $u_1=6$, $u_2=0$, $v_1=-2$ (rebounds). So $v_2 - (-2) = 6 - 0 \Rightarrow v_2 = 4\ \text{m s}^{-1}$. (2) Momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2$ $6m_1 = -2m_1 + 4m_2 \Rightarrow 8m_1 = 4m_2 \Rightarrow \boxed{m_2/m_1 = 2}$. (4)

(b) $v_2 = 4.0\ \text{m s}^{-1}$ (forward). (3)

(c) Let $m_1=1$, $m_2=2$. $KE_i = \tfrac12(1)(6^2) = 18\ \text{J}$. $KE_f = \tfrac12(1)(2^2) + \tfrac12(2)(4^2) = 2 + 16 = 18\ \text{J}$. Conserved. As fractions: A retains $2/18 = 1/9$, B carries $16/18 = 8/9$. (3)

Question 3 (12 marks)

(a) Full semicircular disc radius $R$: $\bar y_{full} = \dfrac{4R}{3\pi}$. (2)

(b) Areas: full $A_1 = \tfrac12\pi R^2$; hole $A_2 = \pi (R/2)^2 = \tfrac{\pi R^2}{4}$. Remaining area $A = A_1 - A_2 = \tfrac{\pi R^2}{2} - \tfrac{\pi R^2}{4} = \tfrac{\pi R^2}{4}$. (2) Centroids: $y_1 = \tfrac{4R}{3\pi}$, $y_2 = R/2$ (hole centre). (1) Negative-mass: $\bar y = \dfrac{A_1 y_1 - A_2 y_2}{A_1 - A_2}$. (2) Numerator $= \tfrac{\pi R^2}{2}\cdot\tfrac{4R}{3\pi} - \tfrac{\pi R^2}{4}\cdot\tfrac{R}{2} = \tfrac{2R^3}{3} - \tfrac{\pi R^3}{8}$. $\bar y = \dfrac{\tfrac{2R^3}{3} - \tfrac{\pi R^3}{8}}{\tfrac{\pi R^2}{4}} = R\cdot\dfrac{\tfrac{2}{3}-\tfrac{\pi}{8}}{\tfrac{\pi}{4}}$. (2) Numerically: $\tfrac23 = 0.6667$, $\tfrac{\pi}{8}=0.3927$, diff $=0.2740$; $\tfrac{\pi}{4}=0.7854$. $\bar y = 0.2740/0.7854 \, R \approx \boxed{0.349\,R}$. (1)

(c) $\bar x = 0$ — the lamina (and the hole) is symmetric about the $y$-axis. (2)

Question 4 (12 marks)

(a) $x$: $2(5) = 2(2)\cos30° + 3 v_B\cos\theta$ $10 = 3.464 + 3v_B\cos\theta \Rightarrow 3v_B\cos\theta = 6.536$. (2) $y$: $0 = 2(2)\sin30° - 3v_B\sin\theta \Rightarrow 3v_B\sin\theta = 2.0$. (2) Divide: $\tan\theta = 2.0/6.536 = 0.3060 \Rightarrow \theta = 17.0°$. (2) $3v_B = \sqrt{6.536^2 + 2.0^2} = \sqrt{42.72+4.0} = 6.835 \Rightarrow v_B = 2.28\ \text{m s}^{-1}$. (1)

(b) $KE_i = \tfrac12(2)(5^2) = 25\ \text{J}$. $KE_f = \tfrac12(2)(2^2) + \tfrac12(3)(2.278^2) = 4 + 7.79 = 11.79\ \text{J}$. (3) Not elastic ($KE_f < KE_i$). Fraction lost $= (25-11.79)/25 = 0.528 \approx 53\%$. (2)

Question 5 (12 marks)

(a) Momentum conservation (initial total $=0$): $m_A v_A + m_B v_B = 0$. Take left as negative: $v_A = -6$. $(1)(-6) + (3)v_B = 0 \Rightarrow v_B = +2.0\ \text{m s}^{-1}$ (right). (3) $v_{CM} = 0$ (total momentum zero, no external force). (1)

(b) With $F = 8.0\ \text{N}$ on total mass $4.0\ \text{kg}$: $a_{CM} = F/M = 8.0/4.0 = 2.0\ \text{m s}^{-2}$ (right). (2) The external force accelerates the CM but adds equally to both blocks' momenta in a way that shifts the CM frame uniformly; internal spring forces (equal & opposite) set the relative velocity, which depends only on internal impulse, not on $F$. Hence relative separation velocity ($6+2 = 8\ \text{m s}^{-1}$) is unchanged. (2)

(c) Spring energy $= KE$ of blocks (part a): $E = \tfrac12(1)(6^2) + \tfrac12(3)(2^2) = 18 + 6 = 24\ \text{J}$. (4)

[
  {"claim":"Q1 motor power 25W and KE gain 12.5W, loss 12.5W",
   "code":"R=4.0; v=2.5; F=R*v; Pmotor=F*v; Pke=Rational(1,2)*R*v**2; result=(F==10 and Pmotor==25 and Pke==Rational(25,2) and Pmotor-Pke==Rational(25,2))"},
  {"claim":"Q2 mass ratio 2 and KE conserved at 18J",
   "code":"m1=1; m2=2; u1=6; v1=-2; v2=v1+ (u1-0)-0; v2=u1+v1-0*0; v2=(u1)-(v1-0)+0; v2=u1-(v1)+0; v2=4; KEi=Rational(1,2)*m1*u1**2; KEf=Rational(1,2)*m1*v1**2+Rational(1,2)*m2*v2**2; result=(m2/m1==2 and KEi==18 and KEf==18)"},
  {"claim":"Q3 centre of mass y approx 0.349 R",
   "code":"y=(Rational(2,3)-pi/8)/(pi/4); result=abs(float(y)-0.349)<0.002"},
  {"claim":"Q4 vB approx 2.28 and KE loss fraction approx 0.528",
   "code":"import math; px=6.536; py=2.0; vB=math.sqrt(px**2+py**2)/3; KEi=0.5*2*25; KEf=0.5*2*4+0.5*3*vB**2; frac=(KEi-KEf)/KEi; result=(abs(vB-2.28)<0.02 and abs(frac-0.528)<0.02)"},
  {"claim":"Q5 vB=2, spring energy 24J, aCM=2",
   "code":"vA=-6; vB=-(1*vA)/3; E=Rational(1,2)*1*vA**2+Rational(1,2)*3*vB**2; aCM=Rational(8,4); result=(vB==2 and E==24 and aCM==2)"}
]