A particle P is sliding down a frictionless h | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q

Vedclass · Accepted Answer

${t_P} < {t_Q}$

Vedclass · Answer

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q} = \frac{AB}{v}$.For particle $P$, the horizontal component of velocity $v_x$ is given by $v_x = v \cos 	heta$, where $	heta$ is the angle the velocity vector makes with the horizontal. As the particle slides down the bowl, it gains speed due to gravity. The horizontal component of its velocity $v_x$ at any point is greater than or equal to the initial horizontal component $v$ because the particle accelerates as it moves towards the lowest point $C$ and then decelerates, but its speed remains higher than the initial speed $v$ throughout the path until it reaches $B$. Since the horizontal component of velocity for $P$ is always greater than or equal to $v$ throughout the motion, the average horizontal velocity of $P$ is greater than $v$. Therefore, the time taken by $P$ to cover the same horizontal distance $AB$ is less than the time taken by $Q$. Thus, ${t_P} < {t_Q}$.

Similar Questions

$A$ small bucket of mass $M \, kg$ is attached to a non-extensible string of length $L \, m$. The bucket is released from rest when the string is in a horizontal position. At its lowest point,the bucket scoops up $m \, kg$ of water and rises to a height $h$. The height $h$ (in meters) is:

$A$ machine which is $75\%$ efficient, uses $12 \, J$ of energy in lifting up a $1 \, kg$ mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in $m \, s^{-1}$):

$A$ car of mass $1000 \ kg$ has a motor of efficiency $20 \%$. If burning of one litre of petrol supplies $6 \times 10^7 \ J$ of energy,the amount of petrol used in accelerating the car from rest to $43.2 \ km \ h^{-1}$ is (in $cc$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module