Question

A shell is shot with an initial velocity ₀ of 20 m/s, at an angle of θ₀ = 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass . 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 andthat air drag is negligible?

Answer 1

Concepts and reason

The concepts used to solve this problem are projectile motion and the law of conservation of momentum.

Initially, use the expression in terms of initial velocity and the angle made by the initial velocity with the horizontal to determine the horizontal component of the initial velocity of the shell.

Finally, use the law of conservation of momentum to determine the speed of second fragment immediately after the explosion.

Fundamentals

The horizontal component of the initial velocity in a projectile motion is given as follows:

Here, the horizontal component of the initial velocity is , the initial velocity is , and the angle made by the initial velocity with the horizontal is .

According to the law of conservation of momentum, “the total momentum of an isolated system remains the same”.

The momentum of a body is given by the expression as follows:

$P=mv $

Here, the momentum is , mass of the body is , and its velocity is .

The horizontal component of the initial velocity in a projectile motion is given as follows:

Substitute for and for .

According to the law of conservation of momentum, “the total momentum of an isolated system remains the same”.

The initial momentum of the shell at the top of the trajectory before explosion is given as follows:

$P=mv $

Here, the initial momentum is .

The shell explodes into two fragments of equal mass at the top of the trajectory.

The masses of the two fragments will be half of the mass of the shell.

Thus:

Here, the mass of the first and second fragments are and , respectively.

The final momentum at the top of the trajectory after explosion is given as follows:

Here, the final momentum is , the velocity of the first and second fragments are and , respectively.

Applying law of conservation of momentum for the given system:

Substitute for and for .

Substitute for and for .

Modify the above expression.

Rearrange the above expression to obtain the speed of the second fragment.

Substitute for and for .

Ans:

The speed of the second fragment is .