Momentum & Collisions

Q1. (3 marks)

• Linear momentum = product of mass and velocity, $p = mv$; a vector quantity. (1)
• SI unit: $\text{kg·m/s}$ (or $\text{N·s}$). (1)
• $p = 0.20 \times 15 = 3.0\ \text{kg·m/s}$. (1)

Q2. (4 marks)

• Start: $F = \dfrac{dp}{dt}$. (1)
• Rearrange: $F\,dt = dp$. (1)
• Integrate over time interval: $\int_{t_1}^{t_2} F\,dt = \int_{p_1}^{p_2} dp$. (1)
• $J = \int F\,dt = p_2 - p_1 = \Delta p$. Impulse equals change in momentum. (1)

Q3. (4 marks)

• Impulse $J = F\,t = 50 \times 0.40 = 20\ \text{N·s}$. (2)
• $J = \Delta p = mv - 0 \Rightarrow v = J/m = 20/2.0 = 10\ \text{m/s}$. (2)

Q4. (5 marks)

• Two particles exert forces on each other: $F_{12}$ on 1 by 2, $F_{21}$ on 2 by 1. (1)
• Newton's third law: $F_{12} = -F_{21}$. (1)
• $\dfrac{dp_1}{dt} = F_{12}$, $\dfrac{dp_2}{dt} = F_{21}$. (1)
• Add: $\dfrac{d(p_1+p_2)}{dt} = F_{12}+F_{21} = 0$. (1)
• Therefore $p_1 + p_2 = \text{constant}$ (total momentum conserved). (1)

Q5. (3 marks)

• Total momentum is conserved if the net external force on the system is zero. (1.5)
• Or, along a particular direction, if the net external force component in that direction is zero (component conserved). Internal forces do not affect total momentum. (1.5)

Q6. (5 marks)

• (a) Momentum conservation: $3.0(4.0) + 0 = (3.0+1.0)v \Rightarrow 12 = 4v \Rightarrow v = 3.0\ \text{m/s}$. (2)
• (b) $KE_i = \tfrac12(3.0)(4.0)^2 = 24\ \text{J}$. (1)
• $KE_f = \tfrac12(4.0)(3.0)^2 = 18\ \text{J}$. (1)
• Loss $= 24 - 18 = 6.0\ \text{J}$. (1)

Q7. (4 marks)

• Elastic 1D, equal masses: $v_1 = u_1$ formula gives $v_1 = 0$, $v_2 = u$. (2)
• So the moving mass stops and the target moves off with speed $u$. (1)
• Comment: equal-mass elastic collision results in exchange of velocities. (1)

Q8. (4 marks)

• $e$ = ratio of relative speed of separation to relative speed of approach; for a bounce off ground $e = \sqrt{h_{rebound}/h_{drop}}$. (2)
• $e = \sqrt{0.80/1.8} = \sqrt{0.4444} = 0.667$. (2)

Q9. (4 marks)

• Linear density $\lambda = M/L$; element $dm = \lambda\,dx$. (1)
• $x_{cm} = \dfrac{1}{M}\int_0^L x\,\lambda\,dx = \dfrac{\lambda}{M}\cdot\dfrac{L^2}{2}$. (1)
• Sub $\lambda = M/L$: $x_{cm} = \dfrac{1}{L}\cdot\dfrac{L^2}{2}$. (1)
• $x_{cm} = L/2$ (midpoint). (1)

Q10. (4 marks)

• Total momentum before $= 4 \times 6 = 24\ \text{kg·m/s}$. (1)
• $1 \times 18 + 3 \times v = 24$. (2)
• $18 + 3v = 24 \Rightarrow v = 2\ \text{m/s}$ (forward). (1)

[
{"claim":"Q3 final speed is 10 m/s","code":"J=50*0.40; v=J/2.0; result=(v==10.0)"},
{"claim":"Q6 common velocity 3 m/s and KE loss 6 J","code":"v=(3.0*4.0)/(3.0+1.0); KEi=Rational(1,2)*3*4**2; KEf=Rational(1,2)*4*v**2; result=(v==3.0 and KEi-KEf==6)"},
{"claim":"Q8 coefficient of restitution approx 0.667","code":"e=sqrt(Rational(8,18)); result=(abs(float(e)-0.6667)<0.001)"},
{"claim":"Q10 3kg fragment velocity is 2 m/s","code":"v=symbols('v'); sol=solve(1*18+3*v-4*6,v); result=(sol[0]==2)"}
]