# Newton's Laws of Motion

## Problems from IIT JEE

Problem (IIT JEE 2007): A particle moves in the $x\text{-}y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)= A \left(\hat{\imath} \cos kt - \hat{\jmath}\sin kt\right)$, where $A$ and $k$ are constants. The angle between the force and the momentum is,

1. 0 degree
2. 30 degree
3. 45 degree
4. 90 degree

Solution: Given, linear momentum of the particle is, \begin{align} \vec{p}=A(\hat\imath\cos kt-\hat\jmath \sin kt). \end{align} By Newton's second law, the force on the particle is, \begin{align} \vec{F}=\mathrm{d}\vec{p}/\mathrm{d}t=A(-\hat\imath\,k\sin kt-\hat\jmath\,k\cos kt). \end{align} The cosine of the angle between $\vec{p}$ and $\vec{F}$ is given by, \begin{align} \cos\theta=\frac{\vec{p}\cdot\vec{F}}{|\vec{p}||\vec{F}|}=-\sin kt\cos kt+\sin kt \cos kt=0, \end{align} and hence $\theta={90}\;\mathrm{degree}$.

Problem: Two particles of mass $m_1$ and $m_2$ in projectile motion have velocities $\vec{v}_1$ and $\vec{v}_2$ respectively at time $t=0$. They collide at time $t_0$. Their velocities becomes ${\vec{v}_1}^\prime$ and ${\vec{v}_2}^\prime$ at time $2t_0$ while still moving in air. The value of $|(m_1{\vec{v}_1}^\prime+m_2{\vec{v}_2}^\prime)-(m_1\vec{v}_1+m_2\vec{v}_2)|$ is

1. zero
2. $(m_1+m_2)gt_0$
3. $2(m_1+m_2)gt_0$
4. $\frac{1}{2}(m_1+m_2)gt_0$

Solution: Consider $m_1$ and $m_2$ together as a system. Gravitational force, $(m_1+m_2)g$, is the only external force acting on the system. Apply Newton's second law, \begin{align} |\vec{F}_\text{ext}|=(m_1+m_2)g=\frac{|\Delta\vec{p}|}{\Delta t}=\frac{|(m_1{\vec{v}_1}^\prime+m_2{\vec{v}_2}^\prime)-(m_1\vec{v}_1+m_2\vec{v}_2)|}{2t_0-0}\nonumber \end{align} to get $|(m_1{\vec{v}_1}^\prime+m_2{\vec{v}_2}^\prime)-(m_1\vec{v}_1+m_2\vec{v}_2)|=2(m_1+m_2)gt_0$. Download png of this solution.

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