Exam

NTA Abhyas

Year

2020

Subject

Physics

Chapter

Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A particle of mass 2 kg moving with a speed of 6 m/s collides...

A particle of mass $\text{2 kg}$ moving with a speed of $\text{6 m/s}$ collides elastically with another particle of mass $\text{4 kg}$ traveling in same direction with a speed of $\text{2 m/s}$. The maximum possible deflection of the $\text{2 kg}$ particle is

Step-by-step solution

Let situations after collision for balls are



According to question $2{v}_{1} + 4{v}_{3} = 20$

$2{v}_{2} = 4{v}_{4}$

$\dfrac{1}{2} \times 2\left({v}_{1}^{2} + {v}_{2}^{2}\right) + \dfrac{1}{2} \times 4\left({v}_{3}^{2} + {v}_{4}^{2}\right) = \dfrac{1}{2} \times 2 \times {\left(6\right)}^{2} + \dfrac{1}{2} \times 4 \times {\left(2\right)}^{2} = 44$

i.e. ${v}_{1}^{2} + {v}_{2}^{2} + 2{v}_{3}^{2} + \dfrac{{v}_{2}^{2}}{2} = 44$

i.e. ${v}_{1}^{2} + \dfrac{3}{2}{v}_{2}^{2} + {\left(\dfrac{10 - {v}_{1}}{2}\right)}^{2}2 = 44$

i.e. $2{v}_{1}^{2} + 3{v}_{2}^{2} + 100 - 20{v}_{1} + {v}_{1}^{2} = 88$

i.e. $3{v}_{1}^{2} + 3{v}_{2}^{2} - 20{v}_{1} + 12 = 0$

i.e. $\dfrac{12}{{v}_{1}^{2}} - \dfrac{20}{{v}_{1}} + \dfrac{3{v}_{2}^{2}}{{v}_{1}^{2}} + 3 = 0$

Quadratic in $\dfrac{1}{v{}_{1}}$ which is real

$∴ D \ge 0 ∴ {\left(- 20\right)}^{2} - 4 \times 12\left(1 + \dfrac{{v}_{2}^{2}}{{v}_{1}^{2}}\right) \times 3 \ge 0$

$∴ 1 + \dfrac{{v}_{2}^{2}}{{v}_{1}^{2}} \le \dfrac{400}{144}$

$\dfrac{{v}_{2}^{2}}{{v}_{1}^{2}} \le \dfrac{256}{144}$

$- \dfrac{16}{12} \le \dfrac{{v}_{2}}{{v}_{1}} \le \dfrac{16}{12} ⇒ k = tan \theta \le \dfrac{4}{3} ⇒ \theta \le {53}^{o}$