Center of Mass Momentum and Collision JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Center of Mass Momentum and Collision, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2009 (Paper 1)
   If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
   1. A.linear momentum of the system does not change in time
   2. B.kinetic energy of the system does not change in time
   3. C.angular momentum of the system does not change in time
   4. D.potential energy of the system does not change in time
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   Answer: (A)

   On a system of particles if, $$\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext} .}=0$$ then $\quad \overrightarrow{\mathbf{P}}_{\text {system }}=$ constant No other conclusions can be drawn.
2. Q2JEE Advanced Adv 2008 (Paper 1)
   Two balls, having linear momenta $\overrightarrow{\mathbf{p}}_1=p \hat{\mathbf{i}}$ and $\overrightarrow{\mathbf{p}}_2=-p \hat{\mathbf{i}}$, undergo a collision in free space. There is no external force acting on the balls. Let $\overrightarrow{\mathbf{p}}_1^{\prime}$ and $\overrightarrow{\mathbf{p}}_2^{\prime}$ be their final momenta. The following option(s) is/are NOT ALLOWED for any non-zero value of $p, a_1, a_2, b_1, b_2, c_1$ and $c_2$
   1. A.$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}$
   2. B.$\overrightarrow{\mathbf{p}}_1^{\prime}=c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=c_2 \hat{\mathbf{k}}$
   3. C.$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}$
   4. D.$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$ $\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}-c_1 \hat{\mathbf{k}}$ $\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$
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   Answer: A,D

   Initial momentum of the system $\overrightarrow{\mathbf{p}_1}+\overrightarrow{\mathbf{p}_2}=0$ $\therefore$ Final momentum $\overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}}$ should also be zero. Option (b) is allowed because if we putc $c_1=-c_2 \neq 0, \overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}}$ will be zero. Similarly, we can check other options. $\therefore$ correct options are (a) and (d).
3. Q3JEE Advanced Adv 2007 (Paper 1)
   Statement I In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Statement II In an elastic collision, the linear momentum of the system is conserved.
   1. A.Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
   2. B.Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I
   3. C.Statement I is true, Statement II is false
   4. D.Statement I is false, Statement II is true
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   Answer: (B)

   In case of elastic collision, coefficient of restitution $e=1$ or relative speed of separation. $\therefore$ Option (b) is correct.