jee-main 2019 Q3

jee-main · India · session1_12jan_shift1 Momentum and Collisions Elastic Collision – Velocity or Mass Determination

A simple pendulum, made of a string of length $l$ and a bob of mass $m$, is released from a small angle $\theta _ { 0 }$. It strikes a block of mass $M$, kept on horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $\theta _ { 1 }$. Then $M$ is given by:
(1) $m \left( \frac { \theta _ { 0 } - \theta _ { 1 } } { \theta _ { 0 } + \theta _ { 1 } } \right)$
(2) $m \left( \frac { \theta _ { 0 } + \theta _ { 1 } } { \theta _ { 0 } - \theta _ { 1 } } \right)$
(3) $\frac { m } { 2 } \left( \frac { \theta _ { 0 } + \theta _ { 1 } } { \theta _ { 0 } - \theta _ { 1 } } \right)$
(4) $\frac { m } { 2 } \left( \frac { \theta _ { 0 } - \theta _ { 1 } } { \theta _ { 0 } + \theta _ { 1 } } \right)$

A simple pendulum, made of a string of length $l$ and a bob of mass $m$, is released from a small angle $\theta _ { 0 }$. It strikes a block of mass $M$, kept on horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $\theta _ { 1 }$. Then $M$ is given by:\\
(1) $m \left( \frac { \theta _ { 0 } - \theta _ { 1 } } { \theta _ { 0 } + \theta _ { 1 } } \right)$\\
(2) $m \left( \frac { \theta _ { 0 } + \theta _ { 1 } } { \theta _ { 0 } - \theta _ { 1 } } \right)$\\
(3) $\frac { m } { 2 } \left( \frac { \theta _ { 0 } + \theta _ { 1 } } { \theta _ { 0 } - \theta _ { 1 } } \right)$\\
(4) $\frac { m } { 2 } \left( \frac { \theta _ { 0 } - \theta _ { 1 } } { \theta _ { 0 } + \theta _ { 1 } } \right)$

Paper Questions

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