You visited us 1 times! Enjoying our articles? Unlock Full Access!

Newton's Law of Gravitation

A small ball ...

Question

A small ball of mass $m$ is projected with a minimum horizontal velocity $v_{0}$ on a smooth wedge of mass $M$ so that it will reach the highest point P of the wedge. The value of $v_{0}$ come out to be $\sqrt{(a + \frac{4 m}{M}) g R}$ . Find the value of $a$ .

Open in App

Solution

If the ball reaches at point P, the velocity of the ball with respect to wedge should be $\sqrt{g R}$ .
Using work-energy theorem from centre of mass frame at A and the highest point P
$W_{e x t} + W_{i n t} = W_{e x t} + 0 = (Δ K)_{c m}$
$W_{g r a v i t y} = (Δ K)_{C M}$
$- m g (2 R) = {[\frac{1}{2} μ v_{r e l}^{2}]}_{f i n a l} - {[\frac{1}{2} μ v_{r e l}^{2}]}_{i n i t i a l}$
$- - m g 2 R = \frac{1}{2} μ (V + v)^{2} - \frac{1}{2} μ v_{0}^{2}$ (i)

$μ = (\frac{m M}{m + M})$ and $v = \sqrt{g R}$
As there is no external forces acting on the system in horizontal direction. The linear momentum of the system should be conserved.
i.e., $m v_{0} = m (V - v) + M V$
$\Rightarrow V = \frac{m (v_{0} + v)}{(m + M)}$ (ii)
From equations (i) and (ii), we get
$\Rightarrow v_{0} = \sqrt{(5 + \frac{4 m}{M}) g R}$

Suggest Corrections

Similar questions

Q. A small ball of mass

m

is projected with a minimum horizontal velocity

v_{0}

on a smooth wedge of mass

M

so that it will reach the highest point

P

of the wedge. Find the value of

v_{0}

Q. A block of mass

m_{1}

is pushed towards the movable wedge of mass

m_{2}

and height

h

, with a velocity

v_{0}

. All surfaces are smooth. The minimum value of

v_{0}

for which the block will reach the top of the wedge is

A small block of mass m is lying at rest at point P of a wedge having a smooth semi circular track of radius R. The minimum value of horizontal acceleration $a_{o}$ of wedge so that mass can just reach the point Q?

Q. A small block of mass

m

is lying at a rest at point

P

of wedge having a smooth semicircular track of radius

R

. The maximum value of horizontal acceleration

a_{0}

of wedge so that mass can just reach the point

Q

Q. A particle of mass 'm' strikes a smooth stationary wedge of mass M with a velocity

v_{0}

, at an angle

θ

with horizontal if the collision is perfectly inelastic, the impulse on the wedge is:

Join BYJU'S Learning Program

A small ball of mass m is projected with a minimum horizontal velocity v0 on a smooth wedge of mass M so that it will reach the highest point P of the wedge. The value of v0 come out to be √(a+4mM)gR. Find the value of a.

A small ball of mass $m$ is projected with a minimum horizontal velocity $v_{0}$ on a smooth wedge of mass $M$ so that it will reach the highest point P of the wedge. The value of $v_{0}$ come out to be $\sqrt{(a + \frac{4 m}{M}) g R}$ . Find the value of $a$ .