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Question: A gun carriage weighs 30 metric tons and carries a gun of mass 5 metric tons. A shell of mass 5 kg l...

A gun carriage weighs 30 metric tons and carries a gun of mass 5 metric tons. A shell of mass 5 kg leaves the gun at 700 m/s when the carriage is at rest. What is the recoil velocity caused in the carriage? Find the recoil distance if the frictional force is 1000 N.
0.1 m/s; 0.175 m
0.2 m/s; 0.275 m
0.3 m/s; 0.175 m
0.4 m/s ; 0.275 m

Explanation

Solution

Use the law of conservation of momentum to determine the recoil velocity of the carriage. To determine the distance travelled by the carriage by the action of frictional force, use the work-energy theorem. The work done by the friction force is equal to the product of magnitude of the frictional force and distance travelled.

Formula Used: The angular momentum of the body of mass m and velocity v is,
P=mvP = mv
The kinetic energy of the body is,
K.E=12mv2K.E = \dfrac{1}{2}m{v^2}

Complete step by step answer: The total mass of the carriage and guns is,
(30+5)metric tons=35×103kg\left( {30 + 5} \right)\,{\text{metric tons}} = 35 \times {10^3}\,kg

We can use law of conservation of angular momentum to determine the recoil velocity of the carriage.
Mvr=mvM{v_r} = mv
vr=mvM{v_r} = \dfrac{{mv}}{M}

Here, M is the mass of carriage, vr{v_r} is the recoil velocity of carriage, m is the mass of shell and v is the velocity of shell.

Substitute 35×103kg35 \times {10^3}\,kg for M, 5 kg for m and 700 m/s for v in the above equation.
vr=(5kg)(700m/s)35×103kg{v_r} = \dfrac{{\left( {5\,kg} \right)\left( {700\,m/s} \right)}}{{35 \times {{10}^3}\,kg}}
vr=3500m/s35×103\Rightarrow {v_r} = \dfrac{{3500\,m/s}}{{35 \times {{10}^3}}}
vr=0.1m/s\Rightarrow {v_r} = 0.1\,m/s

Therefore, the recoil velocity of carriage is 0.1 m/s.

The work done by the frictional force is,
W=FdW = Fd

Here, F is the frictional force and d is the recoil distance.

According to work-energy relation, the kinetic energy goes into work done by the frictional force.
12Mvr2=Fd\dfrac{1}{2}Mv_r^2 = Fd
d=Mvr22F\Rightarrow d = \dfrac{{Mv_r^2}}{{2F}}

Substitute 35×103kg35 \times {10^3}\,kg for M, 0.1 m/s for vr{v_r}, and 1000 N for F in the above equation.
d=(35×103)(0.1)22(1000)\Rightarrow d = \dfrac{{\left( {35 \times {{10}^3}} \right){{\left( {0.1} \right)}^2}}}{{2\left( {1000} \right)}}
d=3502000\Rightarrow d = \dfrac{{350}}{{2000}}
d=0.175m\Rightarrow d = 0.175\,m

Therefore, the distance travelled by the carriage is 0.175 m.

So, the correct answer is option (A).

Note: The velocity of recoil of the gun and velocity of the shell are in opposite directions. Therefore, the recoil velocity should have a negative sign, but we have taken the magnitude of the velocity. Therefore, the negative sign can be ignored.