Question

A car with a gun mounted on it is kept on the horizontal frictionless surface. The total mass of the car, gun and shell is 50 kg. Mass of each shell is 1 kg. If the shell is fired horizontally with relative velocity 100 m/sec with respect to the gun. What is the recoil speed of the car after the second shot?

Answer 1

In order to find a solution for the above numerical, we should understand the concept of conservation of linear momentum. The total momentum is constant for an isolated system.

Complete step by step answer:
The initial velocity of the wheel w.r.t. ground = u
Let us consider the velocity of car after recoil which acts in the opposite direction which is taken as = −v

Therefore,
Total velocity=$u - ( - v)$
Total velocity=$u + v = 100$……. (1)

Now we will apply conservation of momentum to calculate the initial velocity.

${m_1}v + {m_2}u = 0$

$\Rightarrow m( - v) + mu = 0$

$\Rightarrow 49( - v) + u = 0$

$\ \Rightarrow - 49v + u = 0 \\\ \\\ \$
$\Rightarrow u = 49v$……… (2)

By using both the equation 1&2

Equation 1 implies,

$\ 49v + v = 100 \\\ \Rightarrow 50v = 100 \\\ \Rightarrow v = \dfrac{{100}}{{50}} \\\ \Rightarrow v = 2m{s^{ - 1}} \\\ \$

Note: The momentum of the system should be conserved there must be a condition of zero net external force acting on a system. Here conservation of momentum is not only for the individual particles but it is for the whole system. As per the situation the total momentum in the system might increase or decrease.