Question

Two balls of masses $m$ and $2m$ and momenta $4p$ and $2p$ (in the directions shown) collide as shown in figure.


During collision, the value of linear impulse between them is $J$. In terms of $J$ and $p$, which of the following is the correct condition for elastic collision.

• A. $J=\frac{3}{20}p$
  
• B. $J=p$
  
• C. $J=\frac{20}{3}p$
  
• D. $J=\frac{16}{21}p$
  

Answer 1

The correct option is C $J=\frac{20}{3}p$


Given,

Initial momentum of ball of mass $m$, $p_m=4p$

Initial momentum of ball of mass $2m$, $p_{2m}=2p$

So, initial velocity of mass $m$,
$v_1=\frac{4p}{m}$

Initial velocity of mass $2m$, $v_2=\frac{2p}{2m}=\frac{p}{m}$

The directions of the velocities are shown in Fig. (a).

Let the final momentum of the ball of mass $m$ and $2m$ are $p_m^{\prime }$ and $p_{2m}^{\prime }$ respectively.

Since, we know that impulse on a body is the change of its linear momentum.So, we can write,

For ball of mass $m:$

$-J=-p_m^{\prime }-4p$

$\Rightarrow p_m^{\prime }=J-4p$

so, its final velocity,

$v_1^{\prime }=\frac{J-4p}{2m}$

For ball of mass $2m:$

$J=p_{2m}^{\prime }+2p$

$\Rightarrow p_{2m}^{\prime }=J-2p$

so, its final velocity,

$v_2^{\prime }=\frac{J-2p}{m}$


Since the collision is elastic, so the coeffiecint of the restitution will be one for this case
$e=\frac{v{{}_1}^{{}^{\prime }}+v{{}_2}^{{}^{\prime }}}{v_1+v_2}=1$

Substituting the values, we get
$\Rightarrow 1=\frac{\frac{J-4p}{m}+\frac{J-2p}{2m}}{\frac{4p}{m}+\frac{2p}{2m}}$

$\Rightarrow 1=\frac{3J-10p}{10p}$

$\Rightarrow 1=\frac{3J}{10p}-1$

$\Rightarrow \frac{3J}{10p}=2$

$\therefore J=\frac{20}{3}p$

Hence, option (c) is the correct answer.

why this question: This question checks the application of newtons law of restitution.