Solveeit Logo
Login

Question

Question: A bullet-cart system has total mass \[50m\]where ‘m’ is mass of each bullet fired with velocity \[v\...

A bullet-cart system has total mass 50m50mwhere ‘m’ is mass of each bullet fired with velocity v m/sec⁡v\,m/\sec with respect to cart then, find velocity of cart system after second shot.

Explanation

Solution

Initially, the cart system is at rest. Therefore, we can see the initial momentum of the system is zero. After the first shot, apply the law of conservation of momentum for the single bullet moving in the forward direction and cart of remaining bullets. Now do the same for the second shot and find the velocity of the remaining bullet cart system.

Formula used:
p=mvp = mv
Here, p is the momentum, m is the mass and v is the velocity.

Complete step by step answer:
We can see initially the bullet cart system is at rest. After the first shot, one bullet from the cart is fired and the number of bullets remaining in the cart is now 49. Therefore, we can apply law of conservation of momentum after the first as follows,
50m(0 m/s)=mv−49mv150m\left( {0\,m/s} \right) = mv - 49m{v_1}
⇒mv=49mv1\Rightarrow mv = 49m{v_1}
⇒v1=v49\Rightarrow {v_1} = \dfrac{v}{{49}}
Here, m is the mass of one bullet, v is the velocity of that bullet, and v1{v_1} is the recoil velocity of the remaining bullet cart system.
The negative sign in the above equation for velocity of the cart system implies that the cart system moves in the opposite direction as that of the bullet.
We see the initial momentum of the cart system before the second shot is 49mv149m{v_1}. Now we can apply the conservation of momentum after the second shot as follows,
−49mv1=mv−48mv2- 49m{v_1} = mv - 48m{v_2}
Here, v2{v_2} is the velocity of the cart system of the remaining 48 bullets. The negative sign for v2{v_2} implies that the cart of the remaining 48 bullets moves in the opposite direction as that of the bullet.
We can substitute v1=v49{v_1} = \dfrac{v}{{49}} in the above equation.
−49m(v49)=mv−48mv2- 49m\left( {\dfrac{v}{{49}}} \right) = mv - 48m{v_2}
⇒−v=v−48v2\Rightarrow - v = v - 48{v_2}
⇒−2v=48v2\Rightarrow - 2v = 48{v_2}
∴v2=v24\therefore {v_2} = \dfrac{v}{{24}}

Therefore, the velocity of the cart system after the second shot is v24\dfrac{v}{{24}}.

Note:
Students should not ignore the direction of bodies for which the law of conservation is applied. One should assume the positive direction for one of the moving bodies and negative direction for the second body.