Consider a one - dimensional collision that involves a body of mass m1 originally moving in the positive x direction with speed v0 colliding with a second body of mass m2 originally at rest. The collision could be completely inelastic, with the two bodies sticking together, completely elastic, or some where in between. After the collision, m1 moves with velocity v1 while m2 moves with velocity v2. Match the velocities in column I with possible values in column II. Column - I Column - II (A) If m1 > m2, then v1 (P) – 0.5 v0 (B) If m1 > m2, then v2 (Q) 0.2 v0 (C) If m1 < m2 then v1 (R) 0.9 v0 (D) If m1

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

Consider a one - dimensional collision that involves a body of mass m₁ originally moving in the positive x direction with speed v₀ colliding with a second body of mass m₂ originally at rest. The collision could be completely inelastic, with the two bodies sticking together, completely elastic, or some where in between. After the collision, m₁ moves with velocity v₁ while m₂ moves with velocity v₂. Match the velocities in column I with possible values in column II.

Column - I	Column - II
(A) If m₁ > m₂, then v₁	(P) – 0.5 v₀
(B) If m₁ > m₂, then v₂	(Q) 0.2 v₀
(C) If m₁ < m₂ then v₁	(R) 0.9 v₀
(D) If m₁ < m₂ then v₂	(S) 1.5 v₀

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$v_{1} = \frac{(m_{1} - {em}_{2})}{m_{1} + m_{2}} u_{0} = \frac{(1 - e \frac{m_{2}}{m_{1}})}{1 + \frac{m_{2}}{m_{1}}} u_{0} = \frac{(\frac{m_{1}}{m_{2}} - e)}{(\frac{m_{1}}{m_{2}} + 1)}$

If m₁ > m₂

$u_{0} > u_{1} > (\frac{1 - e}{2}) u_{0} \Rightarrow u_{0} > u_{1} > 0$

$(1 + e) > u_{2} > \frac{(1 + e)}{2} \Rightarrow 2 u_{0} > u_{2} > \frac{u_{0}}{2}$

If m₁ < m₂

$- {eu}_{0} < u_{1} < \frac{1 - e}{2} \Rightarrow - u_{0} < u_{1} < \frac{u_{0}}{2}$

$0 < u_{2} < \frac{(1 + e) u_{0}}{2} \Rightarrow 0 < u_{2} < u_{0}$