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pl answer Q 23

Centre of mass of velocity of shell = 0 (initially at rest)

Therefore using conservation od momentum , 0 = 49 ( m x V ) + m x 200

V = -200/49 m/s

which is 200/49 m/s towards left .

Now , coming to the case when second shell is fired

Velocity of shell w.r.t. ground frame = Velocity of shell w.r.t. car + Velocity of car =(200 -200/49) m/s

Let us say velocity of the car is ‘v’ after firing of second shell and towards (-)x axis by observation.Remember, now our system is of 49 m only.

P

_{i}=49 m x(- 200/49 )= -200 m , so we know P

_{f}will be equal to P

_{i}, by momentum conservation.

-48 m v + m(200–200/49) = -200 m

-48 v + 200 -200/49 = -200

-48 v = -200 (2– 1/49)

48 v = 200 (97/49)

v = 200/48 (48 +49)/49

v = 200(1/48 + 1/49 )

Now , the velocity of the car with respect to earth is (200/49 + 200/48)Comparing with 200/48 + 200/49 we see that 49 = 7x

x = 7

Hope this helps.

Regards

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