JEE Main 2014 · 12 Apr · Shift 1 · Physics Q7

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PhysicsSingle correct+4 / -1Question 7 of 90

Two hypothetical planets of masses $m_1$ and $m_2$ 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 $m_1$ is $v_1$ and that of $m_2$ is $v_2$ )

Two shaded spheres m₁ (left) and m₂ (right) a distance d apart, moving towards each other with speeds v₁ and v₂.

v_1=v_2

v_1=m_2\sqrt{\dfrac{2G}{d(m_1+m_2)}}

v_2=m_1\sqrt{\dfrac{2G}{d(m_1+m_2)}}

v_1=m_1\sqrt{\dfrac{2G}{d(m_1+m_2)}}

v_2=m_2\sqrt{\dfrac{2G}{d(m_1+m_2)}}

v_1=m_2\sqrt{\dfrac{2G}{m_1}}

v_2=m_1\sqrt{\dfrac{2G}{m_2}}

Answer and worked solution

Official answer: B

Momentum: $m_1v_1=m_2v_2$ . Energy: $\frac12m_1v_1^2+\frac12m_2v_2^2=\dfrac{Gm_1m_2}d$ .

Substituting $v_2=\frac{m_1}{m_2}v_1$ : $\frac12m_1v_1^2\dfrac{m_1+m_2}{m_2}=\dfrac{Gm_1m_2}d$ , so
$v_1=m_2\sqrt{\frac{2G}{d(m_1+m_2)}},\qquad v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)}}.$

Topics:Gravitational field and potential Conservation of linear momentum

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 7