Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

A shell is shot with an initial velocity ${\vec{v}}_{0}$ of $20 m s^{- 1},$ at an angle of $θ_{0} = 60 °$ with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass in the figure shown below. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

Accepted Answer

Given:

Initial velocity of shell is, ${\vec{v}}_{0} = 20 m s^{- 1}$ .

Angle of projection of shell is, $θ_{0} = 60^{0}$ .

Suppose the mass of the shell is, $m$ .

At the top of the trajectory, the shell explodes into two fragments of equal masses.

Each fragment of mass is $m_{1} = m_{2} = \frac{m}{2}$ .

Velocity of one fragment at the top of the trajectory is $v_{1} = 0$ , as it falls vertically at position, $x_{1} = \frac{R}{2}$ .

Position of second fragment $x_{2} = ?$

(a) Range of shell as it falls at same horizontal level is,

$R = \frac{v^{2} \sin (2 θ_{0})}{g} . . . (1)$ .

Substitute the given values in equation (1)

$\Rightarrow R = \frac{{(20)}^{2} \sin (2 \times 60^{0})}{9.8}$

$\Rightarrow R = 35.35 m$ .

(b) As the shell explodes only due to internal forces, it will not affect the motion of center of mass of the shell. Only force acting on the shell is gravity, i.e., vertically downwards, so, horizontal position of center of mass will not change.

Position of center mass is, $x_{cm} = R$ .

Using the equation of coordinate of center of mass

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} . . . (2)$ .

Substitute the given values in the equation (2)

$R = \frac{(\frac{m}{2} \times \frac{R}{2}) + (\frac{m}{2} \times x_{2})}{\frac{m}{2} + \frac{m}{2}}$

$x_{2} = \frac{3 R}{2} = \frac{3 \times 35.35}{2} = 53.03 m$ .

So, the second fragment lands at $53.03 m$ .

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