A body of mass, m, resting on a smooth inclined plane at an angle $Θ$ to the horizontal is
connected to a mass M over a smooth pulley (Figure 4). Find the velocity, v of the object
of mass m when it has moved a distance b up the plane.

v = \sqrt{\frac{2 g b (M - m s i n Θ)}{M + m}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

v = \sqrt{\frac{g b (M + m s i n Θ)}{M + m}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = \sqrt{\frac{2 g b (M - m s i n Θ)}{M - m}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = \sqrt{\frac{g b (M - m s i n Θ)}{M - m}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $v = \sqrt{\frac{2 g b (M - m s i n Θ)}{M + m}}$
Decrease in PE of system $= M g b m g b s i n Θ$
Increase in KE of system $= 12 (M + m) v^{2}$

Suggest Corrections

Similar questions

Q. A small disc of mass

m

is placed on a body of mass

M

at rest. If the disc is set in horizontal motion with velocity

v

as shown,

(Assume that disc leaves from surface of

M

vertically at A and all surfaces are smooth)
Find velocity of mass

M,

when disc leaves the surface of

M .

A uniform rod of mass 'M' and length 'a' lies on a smooth horizontal plane. A particle of mass m moving at a speed 'v' perpendicular to the length of the rod strikes it at a distance $\frac{a}{4}$ from the center and stops after the collision. Find (a) the linear velocity of the center of the rod and (b) the angular velocity of the rod about its center just after the collision.

Q. As shown in the figure a body of mass