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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Centre of Mass, Linear momentum, Collision

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Centre of Mass, Linear momentum, Collision

Two particles having masses m_1 and m_2 undergo a head-on...

Two particles having masses m1m_1 and m2m_2 undergo a head-on collision. Their initial speeds are u1u_1 and u2u_2, travelling in the same direction. The collision commences at t=0t = 0 and the particles interact over a time interval Δt\Delta t. Throughout the collision, the speed of the first particle changes according to the relation
v(t)=u1+tΔt(v1−u1)v(t) = u_1 + \dfrac{t}{\Delta t}(v_1 - u_1).
Determine the speed of the second particle as a function of time while the collision takes place.

Options

  1. Au2+m2m1tΔt(v1−u1)u_2 + \dfrac{m_2}{m_1}\dfrac{t}{\Delta t}(v_1 - u_1)
  2. Bu2+m1m2tΔt(v1−u1)u_2 + \dfrac{m_1}{m_2}\dfrac{t}{\Delta t}(v_1 - u_1)
  3. Cu2−m1m2tΔt(v1−u1)u_2 - \dfrac{m_1}{m_2}\dfrac{t}{\Delta t}(v_1 - u_1)
  4. Du2−m2m1tΔt(v1−u1)u_2 - \dfrac{m_2}{m_1}\dfrac{t}{\Delta t}(v_1 - u_1)

Correct answer

C. u2−m1m2tΔt(v1−u1)u_2 - \dfrac{m_1}{m_2}\dfrac{t}{\Delta t}(v_1 - u_1)

Step-by-step solution

By the principle of conservation of linear momentum, the total momentum of the system remains constant throughout the collision.

m1u1+m2u2=m1v1(t)+m2v2(t)m_1 u_1 + m_2 u_2 = m_1 v_1(t) + m_2 v_2(t)

Given the velocity of the first particle as a function of time:

v1(t)=u1+tΔt(v1−u1)v_1(t) = u_1 + \dfrac{t}{\Delta t}(v_1 - u_1)

Substituting v1(t)v_1(t) into the momentum conservation equation:

m1u1+m2u2=m1(u1+tΔt(v1−u1))+m2v2(t)m_1 u_1 + m_2 u_2 = m_1 \left( u_1 + \dfrac{t}{\Delta t}(v_1 - u_1) \right) + m_2 v_2(t)

m1u1+m2u2=m1u1+m1tΔt(v1−u1)+m2v2(t)m_1 u_1 + m_2 u_2 = m_1 u_1 + m_1 \dfrac{t}{\Delta t}(v_1 - u_1) + m_2 v_2(t)

Cancelling m1u1m_1 u_1 from both sides:

m2u2=m1tΔt(v1−u1)+m2v2(t)m_2 u_2 = m_1 \dfrac{t}{\Delta t}(v_1 - u_1) + m_2 v_2(t)

Rearranging to solve for v2(t)v_2(t):

m2v2(t)=m2u2−m1tΔt(v1−u1)m_2 v_2(t) = m_2 u_2 - m_1 \dfrac{t}{\Delta t}(v_1 - u_1)

v2(t)=u2−m1m2tΔt(v1−u1)v_2(t) = u_2 - \dfrac{m_1}{m_2} \dfrac{t}{\Delta t}(v_1 - u_1)
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