Question

A particle of mass M1 moving with inertial velocity Vo is incident on a stationary particle of mass M2. After collision, M1 was deflected through an angle ∅ and M2 an angle ∅. If the velocities of the particles after collision were V1 and V2 respectively. Show that for elastic collision;

(a) V2=V2o.M1/M1+M2.cos∅

(b) (V1/Vo)=2(V1/Vo)M1/M1+M2.cosx+M2-M1/M1+M2

(c) M2/M1=sin/sin(∅+x)

Answer 1

$P_1 + P_2 = P_1'+P_2' = M_1V_o$

$P_t = \sqrt{(P_1')^2+(P_2')^2+2P_1'P_2'cos\alpha}$

$\sqrt{(M_1V_1)^2+(M_2V_2)^2+2M_1V_1M_2V_2cos\theta} = M_1V_o$

$(M_1V_1)^2+(M_2V_2)^2+2M_1V_1M_2V_2cos\theta = (M_1V_o)^2$ $/ : M_1^2$

$V_1^2 + (\large\frac{M_2V_2}{M_1})^2$ $+\large\frac{2M_2V_1V_2cos\theta}{M_1} = V_o^2$