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Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration f. The centre of mass has an acceleration 1) 0. 2) f. 3)f/2. 4) 2f. I also want to know whens the does the centre of mass has an acceleration 0, f, f/2, 2f
Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration f. The centre of mass has an acceleration        1) 0.    2) f.          3)f/2.                    4) 2f.                                              I also want to know whens the does the centre of mass has an acceleration 0, f, f/2, 2f


4 years ago

Raman Mishra
67 Points
							Let acm be the acceleration of the centre of mass of the two particles. Then,                $a_{cm}=\frac{m_{1}a_{1}+m_{2}a_{2}}{m_{1}+m_{2}}$Here since the particles are identical i.e, m1=m2=m(say)          $a_{cm}=\frac{m(a_{1}+a_{2})}{2m}=\frac{a_{1}+a_{2}}{2}$               (1)Taking a1=0(particle at rest) and a2= f, we get        acm=f/2   Ans.Also, putting the desired values of acm in eqn(1), we have:acm =0 when a1+a2 = 0 ;acm = f when a1+a2 = 2f;acm = f/2 when a1+a2 = f andacm = 2f when a1+a2 = 4f.Hope you understood dear Sakshi.

4 years ago
Abcd
13 Points
							[5kg]-------------[2kg]V. Com=m1v1 + m2v2 /m1+m2 It is given that mass of 2 kg is at rest.....: V. Com=5×7/7         =5m/sec ...😊

one year ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions