pl answer Q 23
Dear student,
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
Remember, now our system is of 49 m only.
Pi =49 m x(- 200/49 )= -200 m , so we know Pf will be equal to Pi, by momentum conservation.
Comparing with 200/48 + 200/49 we see that 49 = 7x
x = 7
Hope this helps.
Regards
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.
Pi =49 m x(- 200/49 )= -200 m , so we know Pf will be equal to Pi, 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