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5.work,energy,power and collision — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics5.work,energy,power and collision1 Mark
MCQ
A mass $M$ moving with a certain speed $V$ collides elastically with another stationary mass $m$. After the collision, the masses $M$ and $m$ move with speeds $V^{\prime}$ and $v$, respectively. All motion is in one dimension. Then,
  • A
    $V=V^{\prime}+v$
  • B
    $V^{\prime}=V+v$
  • C
    $V^{\prime}=\frac{(V+v)}{2}$
  • ✓
    $v=V+V^{\prime}$
✓

Answer

Correct option: D.
$v=V+V^{\prime}$
d
(d)

Collision is elastic, so both linear momentum and kinetic energy are conserved.

We have following situation,

According to figure,

$M V=M V^{\prime}+m v \ldots$ $(i)$ (linear momentum conservation)

$\frac{1}{2} M V^2=\frac{1}{2} M V^{\prime 2}+\frac{1}{2} m v^2 \dots(ii)$

(kinetic energy conservation)

From Eqs. $(i)$ and $(ii)$, we get

$M\left(V-V^{\prime}\right)=m v \dots(iii)$

and $M\left(V^2-V^{\prime 2}\right)=m v^2 \dots(iv)$

Dividing Eq. $(iv)$ by Eq. $(iii)$, we have

$\frac{M(V^2-V^{\prime 2})}{M(V-V^{\prime})}=\frac{mv^2}{mv}$

$\text { or } V+V^{\prime}=v$

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