Level 1 — RecognitionMomentum & Collisions

Momentum & Collisions

30 marksprintable — key stays hidden on paper

Level 1: Recognition

Time: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each)

Select the single best answer.

Q1. The SI unit of linear momentum is: (a) kg·m/s² (b) kg·m/s (c) kg·m²/s (d) N·m

Q2. The impulse–momentum theorem states that impulse equals: (a) change in kinetic energy (b) rate of change of momentum (c) change in momentum (d) change in force

Q3. Linear momentum of an isolated system is conserved because of: (a) Newton's first law (b) Newton's second law (c) Newton's third law (d) conservation of energy

Q4. In a perfectly inelastic collision, the quantity that is NOT conserved is: (a) total momentum (b) total kinetic energy (c) total mass (d) total charge

Q5. For a perfectly elastic collision, the coefficient of restitution $e$ equals: (a) 0 (b) between 0 and 1 (c) 1 (d) greater than 1

Q6. Two objects of equal mass collide elastically in 1D, one initially at rest. After collision: (a) they move together (b) they exchange velocities (c) both stop (d) both reverse

Q7. The centre of mass of a uniform thin rod of length $L$ lies at a distance from one end: (a) $L/4$ (b) $L/3$ (c) $L/2$ (d) $2L/3$

Q8. The centre of mass of a uniform semicircular arc of radius $R$ lies at a distance from the centre: (a) $\frac{R}{2}$ (b) $\frac{2R}{\pi}$ (c) $\frac{4R}{3\pi}$ (d) $\frac{3R}{4}$

Q9. The acceleration of the centre of mass of a system depends only on: (a) internal forces (b) external forces (c) total kinetic energy (d) coefficient of restitution

Q10. In the rocket equation, thrust arises from: (a) gravity (b) external air pressure (c) the rate at which exhaust mass leaves times its relative velocity (d) the mass of the payload

Section B — True / False with justification (2 marks each: 1 T/F, 1 reason)

Q11. "Momentum is conserved in every collision, elastic or inelastic."

Q12. "In a 2D elastic collision between two equal masses (one at rest), the two final velocity vectors are always perpendicular."

Q13. "If a net external force acts on a system, its total momentum can never be conserved in any direction."

Q14. "The centre of mass of a uniform triangular lamina lies at its centroid, i.e. at $1/3$ the height from the base."

Section C — Matching (1 mark each pair, 6 marks)

Q15. Match each quantity/system in Column I with its correct expression/property in Column II.

Column I	Column II
(i) Impulse	(P) $\frac{4R}{3\pi}$
(ii) Coefficient of restitution	(Q) $\vec{F}\,\Delta t$
(iii) CoM of uniform semicircular disc	(R) $\dfrac{v_2 - v_1}{u_1 - u_2}$
(iv) Maximum KE loss occurs in	(S) perfectly inelastic collision
(v) Rocket equation	(T) $v = u\ln\!\frac{m_0}{m}$
(vi) Momentum	(U) $m\vec{v}$

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) kg·m/s. Since $p=mv$ : units = kg × (m/s).

Q2. (c) change in momentum. From $\vec{J}=\int\vec{F}\,dt=\Delta\vec{p}$ .

Q3. (c) Newton's third law. Internal forces come in equal-and-opposite pairs that cancel, so total $\vec{p}$ stays constant.

Q4. (b) kinetic energy. Momentum, mass, charge conserved; KE lost to deformation/heat.

Q5. (c) 1. Elastic ⇒ no KE loss ⇒ $e=1$ .

Q6. (b) exchange velocities. For $m_1=m_2$ elastic 1D: $v_1=u_2,\ v_2=u_1$ .

Q7. (c) $L/2$ . By symmetry of uniform rod.

Q8. (b) $\frac{2R}{\pi}$ for a semicircular arc. (Disc would be $\frac{4R}{3\pi}$ .)

Q9. (b) external forces. $\vec{a}_{CM}=\vec{F}_{ext}/M$ ; internal forces cancel.

Q10. (c) thrust $=v_{rel}\,\dfrac{dm}{dt}$ .

Q11. TRUE (1). Reason (1): In any collision the internal forces are action–reaction pairs and external impulses are negligible during the short contact, so total momentum is conserved regardless of energy loss.

Q12. TRUE (1). Reason (1): For equal masses with one at rest, momentum and KE conservation give $\vec{u}_1=\vec{v}_1+\vec{v}_2$ and $u_1^2=v_1^2+v_2^2$ ; squaring the first and comparing gives $\vec{v}_1\cdot\vec{v}_2=0$ , i.e. a $90°$ angle (provided neither final velocity is zero).

Q13. FALSE (1). Reason (1): Momentum is conserved along any direction in which the net external force component is zero, even if a net force acts in another direction (e.g. horizontal momentum conserved under gravity).

Q14. TRUE (1). Reason (1): For a uniform triangle, symmetry/integration gives the CoM at the centroid, located at $h/3$ above the base (intersection of medians).

Section C (1 mark per correct pair)

Q15.

(i) Impulse → (Q) $\vec{F}\Delta t$
(ii) Coefficient of restitution → (R) $\frac{v_2-v_1}{u_1-u_2}$
(iii) CoM of semicircular disc → (P) $\frac{4R}{3\pi}$
(iv) Maximum KE loss → (S) perfectly inelastic collision
(v) Rocket equation → (T) $v=u\ln\frac{m_0}{m}$
(vi) Momentum → (U) $m\vec{v}$

[
  {"claim":"Equal-mass elastic 1D collision exchanges velocities: with u1=u,u2=0 -> v1=0,v2=u",
   "code":"m,u=symbols('m u'); v1,v2=symbols('v1 v2'); sol=solve([m*u-(m*v1+m*v2), m*u**2-(m*v1**2+m*v2**2)],[v1,v2],dict=True); ans=[s for s in sol if s[v1]!=u or s[v2]!=0]; result=(ans[0][v1]==0 and ans[0][v2]==u)"},
  {"claim":"2D equal-mass, one at rest: final velocities are perpendicular (dot product zero)",
   "code":"v1x,v1y,v2x,v2y,u=symbols('v1x v1y v2x v2y u'); eqs=[v1x+v2x-u, v1y+v2y, v1x**2+v1y**2+v2x**2+v2y**2-u**2]; expr=(u**2 - (v1x**2+v1y**2) - (v2x**2+v2y**2)); dot=v1x*v2x+v1y*v2y; check=simplify(expr - (-2*dot)); result=(check==0)"},
  {"claim":"Elastic collision means coefficient of restitution e=1 (KE conserved implies e=1)",
   "code":"m1,m2,u1,u2,v1,v2=symbols('m1 m2 u1 u2 v1 v2'); sol=solve([m1*u1+m2*u2-(m1*v1+m2*v2), m1*u1**2+m2*u2**2-(m1*v1**2+m2*v2**2)],[v1,v2],dict=True); s=[d for d in sol if d[v1]!=u1][0]; e=simplify((s[v2]-s[v1])/(u1-u2)); result=(e==1)"},
  {"claim":"CoM of semicircular arc radius R is 2R/pi from centre",
   "code":"R,t=symbols('R t',positive=True); ybar=integrate(R*sin(t)*R,(t,0,pi))/integrate(R,(t,0,pi)); result=(simplify(ybar-2*R/pi)==0)"}
]