Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

A block of mass $m$ is moving with speed $v$ on a wedge of mass $M$ which is placed upon a horizontal surface. Track $A B$ is a circular track of radius $R$ . What is the minimum speed $v$ , so that the block reaches point $A$ (All surfaces are frictionless)?

Accepted Answer

As given, small block of mass $m$ , is moving with velocity $v$ , as shown in figure.
The block follow linear motion till it reaches point B. Let, the velocity of wedge of mass $M$ , when smaller block starts moving be $v_{1}$ . Then, by applying law of conservation of linear momentum for two bodies (wedge-block system), we can write,
$0 = m v - (M + m) v_{1}$
$\Rightarrow v_{1} = \frac{m}{M + m} v . . . (1)$

Now after reaching point B, the block of mass $m$ , follows the circular path of radius $R$ , as shown in figure and moves upward till it reaches point A. As the kinetic energy of the system changes, the energy of the system must be conserved. i.e.,
Total energy of the wedge with block (K.E.+P.E.) = Total energy of the block (K.E.+P.E.)
After substituting values we get,
$\frac{1}{2} (M + m) {v_{1}}^{2} = \frac{1}{2} m v^{2} - m g R$
(As the block goes up, acceleration due to gravity comes in picture)
now substituting value of $v_{1}$ from equation (1), we get,
$\Rightarrow \frac{1}{2} (M + m) {(\frac{m v}{M + m})}^{2} = \frac{1}{2} m v^{2} - m g R$
$\Rightarrow \frac{m^{2} v^{2}}{M + m} = m v^{2} - 2 m g R$
$\Rightarrow \frac{m v^{2}}{M + m} = v^{2} - 2 g R$
$\Rightarrow v^{2} - \frac{m v^{2}}{M + m} = 2 g R$
$\Rightarrow (\frac{M + m - m}{M + m}) v^{2} = 2 g R$
$\Rightarrow v^{2} = 2 g R (\frac{M + m}{M})$
$\Rightarrow v = \sqrt{\frac{2 g R (M + m)}{M}}$ , which is the required velocity of the block.

A block of mass $m$ is moving with speed $v$ on a wedge of mass $M$ which is placed upon a horizontal surface. Track $A B$ is a circular track of radius $R$ . What is the minimum speed $v$ , so that the block reaches point $A$ (All surfaces are frictionless)?

Important Questions on System of Particles and Rotational Motion

Learn and Prepare for any exam you want !

A block of mass m is moving with speed v on a wedge of mass M which is placed upon a horizontal surface. Track AB is a circular track of radius R. What is the minimum speed v, so that the block reaches point A (All surfaces are frictionless)?

Important Questions on System of Particles and Rotational Motion

Learn and Prepare for any exam you want !

A block of mass $m$ is moving with speed $v$ on a wedge of mass $M$ which is placed upon a horizontal surface. Track $A B$ is a circular track of radius $R$ . What is the minimum speed $v$ , so that the block reaches point $A$ (All surfaces are frictionless)?