Momentum & Collisions

Question 1

(a) [6 marks] Momentum: $m_1u_1 = m_1v_1 + m_2v_2$ (1) Energy: $\tfrac12 m_1u_1^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2$ (1 mark each equation = 2)

Rearrange momentum: $m_1(u_1 - v_1) = m_2 v_2$ (i) Rearrange energy: $m_1(u_1^2 - v_1^2) = m_2 v_2^2$ (ii) (1 mark) Divide (ii)/(i): $u_1 + v_1 = v_2$ (1 mark) — this is the key elimination step. Substitute back into (i): $m_1(u_1-v_1) = m_2(u_1+v_1)$, solve: $v_1 = \frac{m_1-m_2}{m_1+m_2}u_1$ (1 mark) Then $v_2 = u_1 + v_1 = \dfrac{2m_1}{m_1+m_2}u_1$. (1 mark)

(b) [5 marks] For general case with both moving, the same divide step gives (with algebra): $u_1 + v_1 = u_2 + v_2$ → rearrange: $v_2 - v_1 = u_1 - u_2$ = $-(u_2-u_1)$. (3 marks for the derivation) Coefficient of restitution $e = \dfrac{v_2 - v_1}{u_1 - u_2} = 1$. (2 marks) Why: elastic ⇒ KE conserved ⇒ relative speed unchanged in magnitude, reversed in direction.

(c) [4 marks] Fraction to target for $m_1$ hitting stationary $m_2$: $f = \frac{4m_1 m_2}{(m_1+m_2)^2}$ (2 marks for formula, derivable from $v_2$) With $m_1=1, m_2=2$: $f = \dfrac{4(1)(2)}{(3)^2} = \dfrac{8}{9} \approx 0.889$. (2 marks)

(d) [7 marks]

def elastic_1d(m1, m2, u1, u2):
    v1 = ((m1 - m2)*u1 + 2*m2*u2) / (m1 + m2)
    v2 = ((m2 - m1)*u2 + 2*m1*u1) / (m1 + m2)
    return v1, v2

(5 marks: correct general formulae; 1 mark clean structure) Test (momentum conservation): (1 mark)

v1, v2 = elastic_1d(2, 3, 5, -1)
assert abs((2*5 + 3*(-1)) - (2*v1 + 3*v2)) < 1e-9

Question 2

(a) [8 marks] Take semicircle in upper half plane, diameter along x-axis. By symmetry $\bar{x}=0$; find $\bar{y}$. (1 mark) Surface density $\sigma$, total mass $M = \sigma \cdot \tfrac12\pi R^2$. (1 mark) Use strips parallel to diameter at height $y$, length $2\sqrt{R^2-y^2}$, thickness $dy$: (2 marks) $\bar{y} = \frac{1}{M}\int_0^R y\,\sigma\,2\sqrt{R^2-y^2}\,dy$ (1 mark) $\int_0^R y\sqrt{R^2-y^2}\,dy = \left[-\tfrac13(R^2-y^2)^{3/2}\right]_0^R = \tfrac13 R^3$. (2 marks) $\bar{y} = \frac{\sigma\cdot 2\cdot \tfrac13 R^3}{\sigma\cdot \tfrac12\pi R^2} = \frac{\tfrac{2}{3}R^3}{\tfrac12\pi R^2} = \frac{4R}{3\pi}.$ (1 mark)

(b) [7 marks] Same density $\rho$. Cylinder: volume $V_c = \pi R^2 \cdot R = \pi R^3$; mass $\propto \pi R^3$. COM at height $R/2$. (1 mark) Hemisphere: volume $V_h = \tfrac23\pi R^3$; COM at $R + \tfrac{3R}{8}$ above base (sits on top of cylinder height $R$). (2 marks) Masses ratio (density cancels): $m_c : m_h = 1 : \tfrac23 = 3:2$. (1 mark) $y_{cm} = \frac{3\cdot \tfrac{R}{2} + 2\cdot\left(R + \tfrac{3R}{8}\right)}{5} = \frac{\tfrac{3R}{2} + 2\cdot\tfrac{11R}{8}}{5} = \frac{\tfrac{3R}{2} + \tfrac{11R}{4}}{5}$ (2 marks) $= \dfrac{\tfrac{6R}{4}+\tfrac{11R}{4}}{5} = \dfrac{17R/4}{5} = \dfrac{17R}{20} = 0.85R.$ (1 mark)

(c) [5 marks] COM fixed because no external horizontal force (frictionless, internal rope forces cancel by Newton's third law ⇒ $a_{cm}=0$, and initially at rest). (2 marks) $m_1 x_1 = m_2 x_2$ with $x_1 + x_2 = 6$. (1 mark) $50 x_1 = 70 x_2 \Rightarrow x_1 = 1.4 x_2$; $1.4x_2 + x_2 = 6 \Rightarrow x_2 = 2.5\ \text{m}$, $x_1 = 3.5\ \text{m}$. (2 marks) (50 kg skater moves 3.5 m, 70 kg skater moves 2.5 m.)

Question 3

(a) [8 marks] Consider system mass $m$ at velocity $v$; in time $dt$ it ejects mass $-dm$ (>0) at exhaust velocity $u$ relative to rocket (backwards). (1 mark) Total momentum conserved incl. external force impulse: $(m)(v) + F_{ext}dt = (m+dm)(v+dv) + (-dm)(v - u)$ (2 marks) Expand, drop 2nd order $dm\,dv$: $F_{ext}dt = m\,dv + u\,dm$ (2 marks) $m\frac{dv}{dt} = -u\frac{dm}{dt} + F_{ext}.$ (1 mark) With $F_{ext}=0$: $dv = -u\,dm/m$; integrate: $v_f - v_i = -u\ln\frac{m_f}{m_i} = u\ln\frac{m_i}{m_f}.$ (2 marks)

(b) [4 marks] $m_i = 2000$, $m_f = 500$ kg. (1 mark) $v_f = 2500\ln(2000/500) = 2500\ln 4$. (1 mark) $\ln 4 = 1.3863$; $v_f = 2500 \times 1.3863 = 3465.7\ \text{m/s}$. (2 marks)

(c) [6 marks] Take direction into wall as positive: $u_i = +15$, $v_f = -10$ m/s. (1 mark) $\Delta p = m(v_f - u_i) = 0.20(-10 - 15) = -5.0\ \text{kg·m/s}$. (2 marks) $F = \Delta p/\Delta t = -5.0/0.02 = -250\ \text{N}$; magnitude $250$ N (directed away from wall). (2 marks) Python: F = abs(0.20*(-10 - 15))/0.02 # = 250.0 (1 mark)

[
  {"claim":"Elastic 1D general formulae conserve momentum for test case", "code":"m1,m2,u1,u2=2,3,5,-1\nv1=((m1-m2)*u1+2*m2*u2)/(m1+m2)\nv2=((m2-m1)*u2+2*m1*u1)/(m1+m2)\nresult = simplify(m1*u1+m2*u2-(m1*v1+m2*v2))==0"},
  {"claim":"KE fraction transferred neutron->deuteron is 8/9", "code":"m1,m2=1,2\nf=Rational(4*m1*m2,(m1+m2)**2)\nresult = f==Rational(8,9)"},
  {"claim":"Composite hemisphere+cylinder COM is 17R/20", "code":"R=symbols('R',positive=True)\nmc,mh=3,2\nycm=(mc*(R/2)+mh*(R+Rational(3,8)*R))/(mc+mh)\nresult = simplify(ycm-Rational(17,20)*R)==0"},
  {"claim":"Semicircle COM integral gives 4R/(3pi)", "code":"R,y=symbols('R y',positive=True)\nnum=integrate(y*2*sqrt(R**2-y**2),(y,0,R))\nden=pi*R**2/2\nresult = simplify(num/den-4*R/(3*pi))==0"},
  {"claim":"Rocket final speed = 2500 ln4", "code":"vf=2500*log(2000/500)\nresult = abs(float(vf)-3465.7)<1.0"}
]