A particle of mass 2 kg moving with a speed of 6 m/s collides elastically with another particle of mass 4 kg…

1. ${37}^0$
2. ${45}^0$
3. ${53}^0$
4. ${60}^0$

Solution

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

$\therefore D\ge 0 \therefore {\left(-20\right)}^2-4\times 12\left(1+\frac{v_2^2}{v_1^2}\right)\times 3\ge 0$

$\therefore 1+\frac{v_2^2}{v_1^2}\le \frac{400}{144}$

$\frac{v_2^2}{v_1^2}\le \frac{256}{144}$

$-\frac{16}{12}\le \frac{v_2}{v_1}\le \frac{16}{12}\Rightarrow k=\tan \theta \le \frac{4}{3}\Rightarrow \theta \le {53}^o$

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