Question

Consider a head-on collision between two particles of masses m1 and m2 The initial speed of the particles are u1 and u2 in the same direction. The collision starts at t = 0 and the particles interact for a time interval Δt. During the collision, the speed of the first particle varies as

v(t) = u1 + (t/Δt)(v1- u1)

Find the speed of the second particle as a function of time during the collision.

Answer 1

Thanks for asking the question!

ANSWER::

Using law of conservation of momentum ,

m₁u₁ + m₂u₂ = m₁v(t) + m₂v'    (v' = speed of second particle during collision)

v' = (m₁/m₂)u₁ + u₂ - (m₁/m₂).v(t)

v' = (m₁/m₂)u₁ + u₂ - (m₁/m₂).{u₁ + (t/Δt)(v₁- u₁)}

v' = u₂ - (m₁/m₂).(t/Δt)(v₁- u₁)

Hope it helps!

Answer 2

Explanation:

As there is no external force acting along horizontal direction

so using the conservation of linear momentum

Pi =Pfm1u1 +m2u2 = m1v(t) +m2v0 (v0 = velocity of second particle during collision)m1u1 +m2u2 =m1(u1+tdt(v1 − u1)) +m2v0m1u1 +m2u2 =m1u1+m1tdt(v1 − u1)+m2v0m2u2 =m1tdt(v1 − u1)+m2v0m2u2 −m1tdt(v1 − u1) =m2v0dividing both sides by m2u2 −m1tm2dt(v1 − u1) =v0.