1. Problem statement. A collision is observed on a frictionless horizontal table between two particles. Particle A has mass $m_A=2$ kg and initial speed $v_{A,i}=5$ m/s directed at $30^{\circ}$ above the +x axis. Particle B has mass $m_B=3$ kg and initial speed $v_{B,i}=2$ m/s directed at $150^{\circ}$ from the +x axis. After the collision particle A is measured to move with speed $v_{A,f}=3$ m/s at $-20^{\circ}$ from the +x axis. Using these measurements, find the final velocity (magnitude and direction) of particle B and compute the change in total kinetic energy to determine if the collision was elastic or inelastic. 2. Principle and formulas used. Momentum conservation in vector form is the primary rule for collisions, because external horizontal forces are negligible. The vector equation is $$m_A\vec{v}_{A,i}+m_B\vec{v}_{B,i}=m_A\vec{v}_{A,f}+m_B\vec{v}_{B,f}$$ In components this gives two scalar equations: $$m_A v_{Ax,i}+m_B v_{Bx,i}=m_A v_{Ax,f}+m_B v_{Bx,f}$$ and $$m_A v_{Ay,i}+m_B v_{By,i}=m_A v_{Ay,f}+m_B v_{By,f}$$ Kinetic energy for each particle is $K=\tfrac{1}{2} m v^{2}$. If total kinetic energy before and after are equal within measurement error the collision is elastic, otherwise it is inelastic. 3. Compute initial velocity components. Compute components of $\vec{v}_{A,i}$ using $\cos$ and $\sin$. $$v_{Ax,i}=v_{A,i}\cos 30^{\circ}=5\times 0.8660254038=4.3301270189\text{ m/s}$$ $$v_{Ay,i}=v_{A,i}\sin 30^{\circ}=5\times 0.5=2.5\text{ m/s}$$ Compute components of $\vec{v}_{B,i}$. $$v_{Bx,i}=v_{B,i}\cos 150^{\circ}=2\times(-0.8660254038)=-1.7320508076\text{ m/s}$$ $$v_{By,i}=v_{B,i}\sin 150^{\circ}=2\times 0.5=1.0\text{ m/s}$$ 4. Compute total initial momentum components. $$p_{x,i}=m_A v_{Ax,i}+m_B v_{Bx,i}=2\times 4.3301270189+3\times(-1.7320508076)=3.4641016151\text{ kg m/s}$$ $$p_{y,i}=m_A v_{Ay,i}+m_B v_{By,i}=2\times 2.5+3\times 1.0=8.0\text{ kg m/s}$$ 5. Compute final momentum of particle A from its measured final velocity. Compute components of $\vec{v}_{A,f}$. $$v_{Ax,f}=v_{A,f}\cos(-20^{\circ})=3\times 0.9396926208=2.8190778624\text{ m/s}$$ $$v_{Ay,f}=v_{A,f}\sin(-20^{\circ})=3\times (-0.3420201433)=-1.0260604299\text{ m/s}$$ Momentum of A after collision is: $$p_{Ax,f}=m_A v_{Ax,f}=2\times 2.8190778624=5.6381557248\text{ kg m/s}$$ $$p_{Ay,f}=m_A v_{Ay,f}=2\times (-1.0260604299)=-2.0521208598\text{ kg m/s}$$ 6. Use momentum conservation to find final momentum (and velocity) of particle B. Final momentum of B is total minus A's final momentum in each component. $$p_{Bx,f}=p_{x,i}-p_{Ax,f}=3.4641016151-5.6381557248=-2.1740541097\text{ kg m/s}$$ $$p_{By,f}=p_{y,i}-p_{Ay,f}=8.0-(-2.0521208598)=10.0521208598\text{ kg m/s}$$ Convert these to velocity components by dividing by $m_B$. $$v_{Bx,f}=\frac{p_{Bx,f}}{m_B}=\frac{-2.1740541097}{3}=-0.7246847032\text{ m/s}$$ $$v_{By,f}=\frac{p_{By,f}}{m_B}=\frac{10.0521208598}{3}=3.3507069533\text{ m/s}$$ 7. Compute magnitude and direction of $\vec{v}_{B,f}$. Magnitude is $$v_{B,f}=\sqrt{v_{Bx,f}^{2}+v_{By,f}^{2}}=\sqrt{(-0.7246847032)^{2}+(3.3507069533)^{2}}\approx 3.428\text{ m/s}$$ Direction (angle from +x) using $\operatorname{atan2}$ gives $$\theta_{B,f}=\operatorname{atan2}(v_{By,f},v_{Bx,f})=\operatorname{atan2}(3.3507069533,-0.7246847032)\approx 102.27^{\circ}$$ This means particle B moves in the second quadrant at about $102.27^{\circ}$ from +x. 8. Check kinetic energy before and after and compute change. Initial total kinetic energy is $$K_i=\tfrac{1}{2}m_A v_{A,i}^{2}+\tfrac{1}{2}m_B v_{B,i}^{2}=\tfrac{1}{2}\times 2\times 5^{2}+\tfrac{1}{2}\times 3\times 2^{2}=25+6=31.0\text{ J}$$ Final total kinetic energy using measured final speeds is $$K_f=\tfrac{1}{2}m_A v_{A,f}^{2}+\tfrac{1}{2}m_B v_{B,f}^{2}=\tfrac{1}{2}\times 2\times 3^{2}+\tfrac{1}{2}\times 3\times (3.428)^{2}$$ $$\quad =9+1.5\times 11.751\approx 9+17.626=26.626\text{ J}$$ Change in kinetic energy is $$\Delta K=K_f-K_i\approx 26.626-31.0=-4.374\text{ J}$$ The negative change means about 4.37 J of kinetic energy was lost to other forms such as heat, sound, or deformation, so the collision is inelastic. 9. Final answer (concise). Final velocity of particle B is approximately $v_{B,f}=3.428$ m/s at angle $102.27^{\circ}$ from the +x axis. The total kinetic energy decreased by about $4.37$ J, so the collision is inelastic.