Question

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 the mass of one shell. If the velocity of the shell with respect to the gun (in its state before firing) is 200 m/s, what is the recoil speed of the car after the second shot? Neglect friction.

Answer 1

Thanks for asking the question!

ANSWER::

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

Therefore , 0 = 49 m x V + m x 200

V = -200/49 m/s

200/49 m/s towards left .

Another shell is fire , then the velocity of the car , with respect to platform is

V' = 200/49 m/s towards left

Another shell is fired , then the velocity of the car with respect to platform is

v' = 200/48 m/s towards left

Now , the velocity of the car with respect to earth is (200/49 + 200/48) = 19400/2352 = 8.24 m/s towards left.

Hope it helps!

Answer 2

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.