Two balls with masses in the ratio of 1 : 2 moving in opposite direction have a head a head - on elastic collision. If their speeds before impact were in the ratio of 3 : 1 respectively, then speeds after impact will have the ratio:

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

Two balls with masses in the ratio of 1 : 2 moving in opposite direction have a head a head - on elastic collision. If their speeds before impact were in the ratio of 3 : 1 respectively, then speeds after impact will have the ratio:

5 : 3

4 : 5

None

7 : 5

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$\frac{m_{1}}{m_{2}} = \frac{1}{2} = \frac{m}{2 m}$

$\frac{v_{1}}{v_{2}} = \frac{3}{1} = \frac{3 v}{v}$

p_i = p_f ⟹ m × 3v – 2m × v = mv₁' + 2mv₂'
v = v₁' + 2v₂' ... (1)

$1 = e = \frac{v_{2} {}^{'}- {v_{1}}^{'}}{4 v} \Rightarrow v_{2} {}^{'}- v_{1} {}^{'}= 4 v$ ..........(2)

(1) + (2) :

$3 v_{2} {}^{'}= 5 v \Rightarrow 5 v \Rightarrow v_{2} {}^{'}= \frac{5 v}{3} - 4 v = \frac{- 7 v}{3}$

$|\frac{v_{1}^{'}}{v_{2}^{'}}| = \frac{7}{5}$ .