A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with a speed v. Friction between the bead and the string may be neglected. Let tp and tq be the respective times taken by P and Q to reach the point B. Then
Time taken by p is
a) less than Q
b) greater than Q
c)equal to Q
A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 along the horizontal string AB, with a speed v. Friction between the bead and the string may be neglected. Let tp and tq be the respective times taken by P and Q to reach the point B. Then
Time taken by p is
a) less than Q
b) greater than Q
c)equal to Q
Dear student,
answer is a.
Assuming there is no gravity here, the speed of P is constant. As particle P moves round the circle the direction of velocity becomes closer to horizontal, so the horizontal component increases. At each point on the hemispherical track the horizontal component of velocity is greater than for particle Q. So P arrives before Q.
(Imagine looking down on both particles P and Q from above. The only motion you would see is the x-component. The 2 particles start with the same x-component of velocity, but that of P increases as the lower curve becomes closer to horizontal. Then, at the midpoint of AB, this x-component of velocity decreases again back to the initial value. In this view, P is moving faster than Q for almost the whole time. It never moves slower than Q. Therefore it arrives quicker at B.)
If there is gravity then the speed of P will increase as it moves toward the lowest point, and decrease again as it rises. So the horizontal component of speed is increasing even faster than in the zero-gravity situation. P will arrive even earlier than in the zero-gravity case.
The question would be more interesting if the initial speed of P (not its horizontal component) is the same as that of Q, and there is gravity.
Dear student,
answer is a.
Assuming there is no gravity here, the speed of P is constant. As particle P moves round the circle the direction of velocity becomes closer to horizontal, so the horizontal component increases. At each point on the hemispherical track the horizontal component of velocity is greater than for particle Q. So P arrives before Q.
(Imagine looking down on both particles P and Q from above. The only motion you would see is the x-component. The 2 particles start with the same x-component of velocity, but that of P increases as the lower curve becomes closer to horizontal. Then, at the midpoint of AB, this x-component of velocity decreases again back to the initial value. In this view, P is moving faster than Q for almost the whole time. It never moves slower than Q. Therefore it arrives quicker at B.)
If there is gravity then the speed of P will increase as it moves toward the lowest point, and decrease again as it rises. So the horizontal component of speed is increasing even faster than in the zero-gravity situation. P will arrive even earlier than in the zero-gravity case.
The question would be more interesting if the initial speed of P (not its horizontal component) is the same as that of Q, and there is gravity.
