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

Byju's Answer

Standard VII

Physics

Law of Conservation of Energy

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. Find the value of $v_{0}$ .

Open in App

Solution

Height of Q from P is $2 R$

Kinetic energy = potential energy

$\frac{1}{2} m v_{o}^{2} \geq m g (2 R)$

$v_{o} \geq 2 \sqrt{g R}$

Hence, minimum horizontal velocity is $2 \sqrt{g R} m s^{- 1}$

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. The value of

v_{0}

come out to be

\sqrt{(a + \frac{4 m}{M}) g R}

. Find the value of

a

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

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:

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

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. Find the value of v0.

Height of Q from P is 2R Kinetic energy = potential energy 12mv2o≥mg(2R) vo≥2√gR Hence, minimum horizontal velocity is 2√gRms−1

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 $v_{0}$ .

Height of Q from P is $2 R$

Kinetic energy = potential energy

$\frac{1}{2} m v_{o}^{2} \geq m g (2 R)$

$v_{o} \geq 2 \sqrt{g R}$

Hence, minimum horizontal velocity is $2 \sqrt{g R} m s^{- 1}$