Solveeit Logo
Login

Question

Question: A cannon of mass \(5m\) (including a shell of mass m) is at rest on a smooth horizontal ground, fire...

A cannon of mass 5m5m (including a shell of mass m) is at rest on a smooth horizontal ground, fires the shell with its barrel at an angle θ\theta with the horizontal at a velocity u relative to the cannon. Find the horizontal distance of the point where the shell strikes the ground from the initial position of the cannon.
A. 16u2sin⁡2θ9g\dfrac{{16{u^2}\sin 2\theta }}{{9g}}
B.u2sin⁡2θ5g\dfrac{{{u^2}\sin 2\theta }}{{5g}}
C.3u2sin⁡2θ5g\dfrac{{3{u^2}\sin 2\theta }}{{5g}}
D.8u2sin⁡2θ5g\dfrac{{8{u^2}\sin 2\theta }}{{5g}}

Explanation

Solution

The initial momentum of the cannon is zero as it is kept at rest. Resolve the components of initial velocity along parallel and perpendicular to the ground. The calculate time for horizontal and vertical velocity separately. Use that information to solve the above question.

Complete step by step answer:
Let,
Mass of the cannon is 5m5m, which is resting on a smooth horizontal ground. The shell is fired at an angle of θ\theta with the horizontal at a velocity u.u.
Now as the initial momentum of the cannon is zero, as it is at rest.
Let the relative velocity of the cannon be u.u.
From momentum of conservation, we have
Hmv=m(u+v)Hmv = m(u + v)
∴V=43,\therefore V = \dfrac{4}{3},
Therefore, the initial velocity=(4+v) = (4 + v)
=4+(4/3).= 4 + (4/3).
=443.= \dfrac{{44}}{3}.
Now along horizontal, we have

∴0=m(u−cos⁡θ−v)−4mv\therefore 0 = m(u - \cos \theta - v) - 4mv
⇒v=4cos⁡θ5\Rightarrow v = \dfrac{{4\cos \theta }}{5}
∴\therefore Velocity of shell along horizontal w.r.t. ground
=4cos⁡θ−ucos⁡θ5= 4\cos \theta - \dfrac{{u\cos \theta }}{5}
=45(ucos⁡θ)= \dfrac{4}{5}(u\cos \theta )
Let xxbe the horizontal displacement.
Time required:
T=2usin⁡θgT = \dfrac{{2u\sin \theta }}{g}
T=(45ucos⁡θ)(2usin⁡θg)T = \left( {\dfrac{4}{5}u\cos \theta } \right)\left( {\dfrac{{2u\sin \theta }}{g}} \right)
⇒T=4u2sin⁡2θ5g\Rightarrow T = \dfrac{{4{u^2}\sin 2\theta }}{{5g}}
Angle of projection=θ= \theta
Range=42sin⁡2θ9g = \dfrac{{{4^2}\sin 2\theta }}{{9g}}
=16u2sin⁡2θ9g= \dfrac{{16{u^2}\sin 2\theta }}{{9g}}

So, the correct answer is “Option A”.

Note:
The time taken by the projectile to fall back on the ground, will be the same along vertical motion as well as along horizontal motion. So the best way to solve this question is to calculate time using the vertical component of velocity. And then use in for the formula of distance for the horizontal motion, to find range. Maximum horizontal distance covered by a projectile if the angle of projection is 450{45^0}.