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 $3 kg$ block $A$ moving with $4 m s^{- 1}$ on a smooth table collides inelastically and head on with an $8 kg$ block $B$ moving with speed $1.5 m s^{- 1}$ towards $A$ . Given, $e = \frac{1}{2}$ .

(a) What are the final velocities of both the blocks?

(b) Find out the impulse of reformation and deformation.

(c) Find out the maximum potential energy of deformation.

(d) Find out loss in kinetic energy of system.

Accepted Answer

Let, $m_{1} = 3 kg$ , $m_{2} = 8 kg$ , $u_{1} = 4 m s^{- 1}$ , $u_{2} = 1.5 m s^{- 1}$ and let final velocities of masses be $v_{1}$ and $v_{2}$ , respectively.

As linear momentum is conserved during collision, we can write,

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v = m_{1} v_{1} + m_{2} v_{2}, \Rightarrow v = \frac{m_{1} u_{1} + m_{2} u_{2}}{(m_{1} + m_{2})} \Rightarrow v = \frac{3 (4) + 8 (1.5)}{3 + 8} = \frac{24}{11} m s^{- 1}$

Also, using, $A$ ,

$\Rightarrow 3 (4) + 8 (1.5) = 3 v_{1} + 8 v_{2} \Rightarrow 3 v_{1} + 8 v_{2} = 24 . . . . (1)$

The coefficient of restitution for the collision can be expressed as,

$e = \frac{velocity of separation after collision}{velocity of approach before collision} = \frac{1}{2}$

$e = \frac{v_{2} - v_{1}}{4 - 1.5} = \frac{1}{2}, \Rightarrow v_{2} - v_{1} = \frac{5}{4} . . . . (2)$

Solving $(1)$ and $(2)$ , we get,

$v_{1} = \frac{14}{11} and v_{2} = \frac{111}{44}$ .

Impulse of reformation is given as change in momentum of any body during reformation.

$J_{R} = |m_{1} (v - u_{1})| = |m_{2} (v - u_{2})|, \Rightarrow J_{R} = |3 (\frac{24}{11} - 4)| = |8 (\frac{24}{11} - \frac{3}{2})| = \frac{60}{11} N s$

Similarly, impulse of deformation is given as change in momentum of any body during deformation.

$J_{D} = |m_{1} (v_{1} - v)| = |m_{2} (v_{2} - v)|, \Rightarrow J_{D} = |3 (\frac{14}{11} - \frac{24}{11})| = |8 (\frac{111}{44} - \frac{24}{11})| = \frac{30}{11} N s$

The potential energy becomes maximum when the kinetic energy is minimum. So, this can occur when the bodies move with the same velocities. It is given as, $\frac{n E}{n + 1}$ where, $n$ is ratio of mass of second and first bodies and $E$ is total initial kinetic energy.

The maximum potential energy of deformation is,

$\frac{(\frac{8}{3}) ((\frac{1}{2}) m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2})}{(\frac{8}{3}) + 1} = \frac{(\frac{8}{3}) ((\frac{1}{2}) 3 {(4)}^{2} + \frac{1}{2} 8 {(1.5)}^{2})}{(\frac{11}{3})} = \frac{8 (24 + 9)}{11} = 24 J$

The loss in kinetic energy of the system is given as,

$∆ K = (\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}) - (\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}), \Rightarrow ∆ K = (\frac{1}{2} 3 (4^{2}) + \frac{1}{2} 8 (2 . 5^{2})) - (\frac{1}{2} 3 {(\frac{14}{11})}^{2} + \frac{1}{2} 8 {(\frac{111}{44})}^{2}) \Rightarrow ∆ K = 33 - 27.88 = 5.11 J$

Important Questions on Centre of Mass, Momentum and Collisions

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