Physics - Center of Mass Question with Solution | TestHub

Initially, the system ($m_1+m_2$) is at rest. By momentum conservation, when $m_1$ moves left at $v_1=6\text{ m/s}$, $m_2$ must move right. $m_1(-v_1)+m_2{v}_2=0\Rightarrow 1(-6)+3v_2=0\Rightarrow v_2=2\text{ m/s}$. The kinetic energy of the system at this instant is $KE=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}(1)(6^2)+\frac{1}{2}(3)(2^2)=18+6=24\text{ J}$. This is the potential energy stored in the spring ($U_{spring}$) at this compressed state. Now, $m_3$ collides elastically with $m_2$. Let $u_2=2\text{ m/s}$ (initial velocity of $m_2$) and $u_3$ (initial velocity of $m_3$). After collision, $m_2$ is at rest ($v_2^{\prime }=0$). For elastic collision: $v_2^{\prime }=\frac{(m_2-m_3)u_2+2m_3{u}_3}{m_2+m_3}=0$. Plugging in values: $\frac{(3-2)(2)+2(2)u_3}{3+2}=0\Rightarrow 2+4u_3=0\Rightarrow u_3=-0.5\text{ m/s}$. So, $m_3$ was moving left at $0.5\text{ m/s}$. The final velocity of $m_3$ is $v_3^{\prime }=\frac{(m_3-m_2)u_3+2m_2{u}_2}{m_2+m_3}=\frac{(2-3)(-0.5)+2(3)(2)}{3+2}=\frac{0.5+12}{5}=2.5\text{ m/s}$ (to the right). After the collision, $m_1$ is still moving left at $6\text{ m/s}$, $m_2$ is at rest. The total momentum of the $(m_1+m_2)$ system is $P=m_1(-6)+m_2(0)=-6\text{ kg m/s}$. The kinetic energy of the $(m_1+m_2)$ system is $K{E}_{post-coll}=\frac{1}{2}(1)(6^2)+\frac{1}{2}(3)(0^2)=18\text{ J}$. The total mechanical energy of the $(m_1+m_2)$ system (KE + stored PE) just after collision is $E_{total}=18\text{ J}+U_{spring}=18+24=42\text{ J}$. When the spring reaches its natural length, its potential energy is zero. Let the final velocities of $m_1$ and $m_2$ be $V_1$ and $V_2$. By momentum conservation for $(m_1+m_2)$ system: $1V_1+3V_2=-6$. By energy conservation: $\frac{1}{2}(1)V_1^2+\frac{1}{2}(3)V_2^2=42$. Substituting $V_1=-6-3V_2$ into the energy equation: $\frac{1}{2}(-6-3V_2{)}^2+\frac{3}{2}{V}_2^2=42$. This simplifies to $(36+36V_2+9V_2^2)+3V_2^2=84\Rightarrow 12V_2^2+36V_2-48=0\Rightarrow V_2^2+3V_2-4=0$. Solving the quadratic equation, $(V_2+4)(V_2-1)=0$, so $V_2=1\text{ m/s}$ or $V_2=-4\text{ m/s}$. If $V_2=1\text{ m/s}$ (to the right), then $V_1=-6-3(1)=-9\text{ m/s}$ (to the left). If $V_2=-4\text{ m/s}$ (to the left), then $V_1=-6-3(-4)=6\text{ m/s}$ (to the right). Since the spring was compressed and $m_1$ was moving left, it will continue to move left as the spring expands, and $m_2$ will move right. Thus, $V_1=-9\text{ m/s}$ and $V_2=1\text{ m/s}$ is the physically correct solution. The final speed of $m_1$ is $|V_1|=9\text{ m/s}$.