The gravitational force between two particles with masses m and M, initially at rest at great separation, pulls them together. When their separation becomes d, then the speed of either particle relative to the other will be
A
√G(m+M)2d
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B
√G(m+M)d
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C
√4G(m+M)d
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D
√2G(m+M)d
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Solution
The correct option is D√2G(m+M)d
Let the speed of mass M and m be v and u respectively when the distance between them reduces to d.
Since external forces are absent, the linear momentum of the system will be conserved.
∴Pi=Pf
⇒0=Mv−mu
⇒u=Mvm....(1)
Also, since external and non-conservative forces are absent, so the mechanical energy of the system will be conserved.
∴Ui+KEi=Uf+KEf
⇒0+0=−GMmd+12Mv2+12mu2
[∵ at a large distance Ui=0 and initial the particles are at rest, hence KEi=0]
from eq. (1),
⇒0=−GMmd+12Mv2+12mM2m2v2
⇒2Gmd=v2+Mmv2=v2(m+M)m
⇒v=√2Gm2(m+M)d=m√2G(m+M)d
putting the value in equation (1),
u=Mm×m√2G(m+M)d
⇒u=M√2G(m+M)d
Since, both particles are moving in opposote direction, so relative speed will be
v−(−u)=v+u=(m+M)√2G(m+M)d
⇒v+u=√2G(m+M)d
Hence, option (d) is correct.
Why this question ?Key Concept - When the external forceon a system is zero, linear momentum of the system is conserved.