Question

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

• A. $37^0$
• B. $45^0$
• C. $53^0$
• D. $60^0$

Answer 1

Answer: (C)

Let situations after collision for balls are

According to question $2v_1+4v_3=20$
$2v_2=4v_4$
$\frac{1}{2}\times 2\left(v_1^2+v_2^2\right)+\frac{1}{2}\times 4\left(v_3^2+v_4^2\right)=\frac{1}{2}\times 2\times {\left(6\right)}^2+\frac{1}{2}\times 4\times {\left(2\right)}^2=44$
i.e. $v_1^2+v_2^2+2v_3^2+\frac{v_2^2}{2}=44$
i.e. $v_1^2+\frac{3}{2}{v}_2^2+{\left(\frac{10-v_1}{2}\right)}^{2}2=44$
i.e. $2v_1^2+3v_2^2+100-20v_1+v_1^2=88$
i.e. $3v_1^2+3v_2^2-20v_1+12=0$
i.e. $\frac{12}{v_1^2}-\frac{20}{v_1}+\frac{3v_2^2}{v_1^2}+3=0$
Quadratic in $\frac{1}{v_1}$ which is real
$\therefore D\ge 0\therefore {\left(-20\right)}^2-4\times 12\left(1+\frac{v_2^2}{v_1^2}\right)\times 3\ge 0$
$\therefore 1+\frac{v_2^2}{v_1^2}\le \frac{400}{144}$
$\frac{v_2^2}{v_1^2}\le \frac{256}{144}$
$-\frac{16}{12}\le \frac{v_2}{v_1}\le \frac{16}{12}\Rightarrow k=tan\theta \le \frac{4}{3}\Rightarrow \theta \le 53^o$