Question

A particle $\text{P}$ is sliding down a frictionless hemispherical bowl. It passes the point $\text{A}$ at $t=0$. At this instant of time, the horizontal component of its velocity is $v$. $\text{A}$ bead $\text{Q}$ of the same mass as $\text{P}$ is ejected from $\text{A}$ at $t=0$ along the horizontal string $\text{AB}$ (see figure) with the speed $v$. Friction between the bead and the string may be neglected. Let $t_P$ and $t_Q$ be the respective time taken by $\text{P}$ and $\text{Q}$ to reach the point $\text{B}$ . Then

• A. $t_P<t_Q$
  
• B. $t_P=t_Q$
  
• C. $t_P>t_Q$
• D. All of these
  

Answer 1

The correct option is A $t_P<t_Q$


For particle $\text{P}$ , motion between $\text{A}$ and $\text{C}$ will be an accelerated one, while between $\text{C}$ and $\text{B}$ a retarded one. But in any case, horizontal component of it’s velocity will be greater than or equal to v. On the other hand in case of particle $\text{Q}$ , it is always equal to $v$. Horizontal displacement of both the particles are equal. So, $t_P<t_Q$

Why this question? Note: For displacement in horizontal direction, we only need to know about the particle's velocity in horizontal direction throughout the journey and the time taken.