Question

namstey ☺
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A gun is mounted on a railroad car . the mass of the car , the gun , the shell , and the operator is 50m where m is the mass one shell . if the velocity of the shell with respect to the gun (in its state before firing ) is 200m/s what is the recoil speed of the car after the second shot ?

Answer 1

$\star \huge \bold{hello}$

$\bold{ \: solution}$:-

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Given :- A car is mounted on a railroad car . the mass of the car , the gun , the shell , and the operator is 50m . and m is the mass of one shell . velocity of muzzle of the shell is 200m/s

initial Vcm = 0

0 = 49 *m * V + m*200

V = - 200/49 m/s

•°• 200/49 m/s towards left side .

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

V ' = 200/49 towards left

* when another shell is fired then then the velocity of car with respect to the platform is

v'' = 200/48 m/s towards left

velocity of car w.r.t the earth is (200/49 + 200/48) m/s towards left .

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$\red{hope \: it \: helps \: you}$

$\star \pink{muskraj} \star$

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.