A gun is mounted on a railroad car . The mass of the car, the gun, the shells and the operator is 50m where m is mass of 1 shell. If the velocity of shell w.r.t gun is 200m/s. What is the recoil speed of car after the second shot. Neglect friction.
Answers
thanks
Answer:I have assumed here (+) x axis to be our (+) direction throughout the solution.
Let us assume that when one shell is fired car gets a velocity ‘u’. By observation, we know that ‘u’ will be directed opposite to (+)x direction ,if the shell is fired in (+)x direction
We know that P i=0; therefore by momentum conservation we kow that P f =0
200 m - 49 mu=0
u=200/49 m/s
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 )
where v is the recoil speed of the car.