Question

granity
accelaration
lokg
n
velocity of 3kg
Block?
349​

Answer 1

Answer:

First of all, find velocity of center of mass of system of bodies.  

We know, velocity of center of mass of system of bodies is given by,

v  

cm

​  

=  

m  

1

​  

+m  

2

​  

 

m  

1

​  

v  

1

​  

+m  

2

​  

v  

2

​  

 

​  

 

Here, m  

1

​  

=2 kg,m  

2

​  

=3 kg

v  

1

​  

=5 m/s        v  

2

​  

=−5 m/s

so, v  

cm

​  

=  

(2+3)

(2×5+3×(−5))

​  

=−1 m/s

Now, velocity of m  

1

​  

 with respect to centre of mass, v  

cm

​  

=5−(−1)=6 m/s

Distance between two bodies is r=1 m.

Dis. b/w m  

1

​  

 and position of centre of mass, r  

1

​  

=  

m  

1

​  

+m  

2

​  

 

m  

2

​  

×r

​  

 

                                                                               =3/5×1=0.6 m

Tension of string =  

r  

1

​  

 

mv  

1 cm

2

​  

 

​  

=  

0.6

2×6  

2

 

​  

=120 N

Explanation: