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 neutron is scattered through (

\equiv

deviation from its original direction)

θ

degree in an elastic collision with an initially stationary deuteron. If the neutron loses

\frac{2}{3}

of its initial

K . E .

to the deuteron then find the value of

θ .

(In atomic mass unit, the mass of a neutron is

1 u

and mass of a deuteron is

2 u

).

Accepted Answer

Given,
Angle of scattering of neutron is, $θ$
Kinetic energy of neutron finally, ${(K_{f})}_{n} = {(K_{i})}_{n} - \frac{2}{3} {(K_{i})}_{n} = \frac{1}{3} {(K_{i})}_{n}$

Let us assume,
Mass of neutron is, $m$
Mass of deuteron is, $4 m$
Initial velocity of neutron is, $u$
Final velocity of neutron is, $v_{1}$
Final velocity of deuteron is, $v_{2}$
Scattering angle of deuteron is, $ϕ$

The above case before and after collision can be shown in below figure,

Collision is elastic, so kinetic energy must be conserved, so, final kinetic energy of deuteron is, ${(K_{f})}_{d} = \frac{2}{3} {(K_{i})}_{n}$ , as ${(\frac{1}{3})}^{r d}$ of kinetic energy of neutron is acquired by neutron itself.
According to question,
$\Rightarrow \frac{1}{2} m {v_{1}}^{2} = \frac{1}{3} \times \frac{1}{2} m u^{2}$
$\Rightarrow {v_{1}}^{2} = \frac{1}{3} u^{2}$
$\Rightarrow v_{1} = \frac{u}{\sqrt{3}}$
Similarly, for deuteron,
$\Rightarrow \frac{1}{2} \times 2 m \times {v_{2}}^{2} = \frac{2}{3} \times \frac{1}{2} m u^{2}$
$\Rightarrow {v_{2}}^{2} = \frac{u^{2}}{3}$
$\Rightarrow v_{2} = \frac{u}{\sqrt{3}}$

On applying Conservation of linear momentum in horizontal direction, we have
$m u = m v_{1} \cos (θ) + 2 m v_{2} \cos (ϕ)$
Putting values of $v_{1} & v_{2}$ , we have
$\Rightarrow u = \frac{u}{\sqrt{3}} \cos (θ) + 2 \frac{u}{\sqrt{3}} \cos (ϕ)$
$\Rightarrow \cos (θ) + 2 \cos (ϕ) = \sqrt{3} . . . (1)$

On applying Conservation of linear momentum in vertical direction, we have
$m v_{1} \sin (θ) = 2 m v_{2} \sin (ϕ)$
$\Rightarrow \sin (θ) = 2 \sin (ϕ), (∵ v_{1} = v_{2} = \frac{u}{\sqrt{3}})$
$\Rightarrow \sqrt{1 - \cos^{2} (θ)} = 2 \sqrt{1 - \cos^{2} (ϕ)}$
$\Rightarrow 1 - \cos^{2} (θ) = 4 (1 - \cos^{2} (ϕ))$
$\Rightarrow 1 - \cos^{2} (θ) = 4 - 4 \cos^{2} (ϕ)$
$\Rightarrow \cos^{2} (θ) - 4 \cos^{2} (ϕ) = - 3$
$\Rightarrow (\cos (θ) + 2 \cos (ϕ)) \times (\cos (θ) - 2 \cos (ϕ)) = - 3$
$\Rightarrow \cos (θ) - 2 \cos (ϕ) = - \sqrt{3} . . . (2)$ , from $(1)$

On adding equation $(1) & (2)$ , we have
$\Rightarrow 2 \cos (θ) = 0$
$\Rightarrow θ = 90 °$

Important Questions on Centre of Mass, Momentum and Collisions

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