Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Gravitational Potential & Energy›Question 29

JEE Main 2014 (12 Apr, Shift 1) · Gravitational Potential & Energy

Gravitational Potential & Energy question from JEE Main 2014 (12 Apr, Shift 1)

Q29JEE Main 2014 · 12 APR · SHIFT 1Difficult

Gravitational Potential & Energy

Figure

Two hypothetical planets of masses m1 and m2 are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is 'd'?

(Speed of m1 is vv1 and that of m2 is vv2)

Show answer and solution

Answer: (b)

  1. Momentum conservation (start from rest): m1v1=m2v2m_1v_1 = m_2v_2
  2. Energy: 12m1v12+12m2v22=Gm1m2d\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{Gm_1m_2}{d}
  3. Put v2=m1v1m2v_2 = \frac{m_1v_1}{m_2}: 12m1v12⋅m1+m2m2=Gm1m2d\frac{1}{2}m_1v_1^2\cdot\frac{m_1+m_2}{m_2} = \frac{Gm_1m_2}{d}
  4. v1=m22Gd(m1+m2)v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)}} and similarly v2=m12Gd(m1+m2)v_2 = m_1\sqrt{\frac{2G}{d(m_1+m_2)}}
  5. So the answer is (b).
Report a mistake

More questions on Gravitational Potential & Energy