Physics question

Question

A bomb initially at rest on a smooth, horizontal surface is exploded into three pieces. Two pieces fly off horizontally at 55.4

angle to each other, a 2.50 kg piece at 22.0 m/s and a 3.20 kg piece at 15.2 m/s. The third piece flies off horizontally at 33.5

m/s. Note: all pieces fly off along the horizontal plane. Determine the following:

(a) The direction of motion of the third piece.

(b) The mass of the third piece.

(c) Explain how you would determine the efficiency of the explosion. Please explain. Note: a final numerical answer is not

required.

Answer 1

In order to talk about direction, let's set up a coordinate system.

Define x and y-axis in the horizontal plane.

Define them so that the original position of the bomb becomes the origin of the two axes.

We also have the freedom to orient the axes in a convenient way --

let's make the positive x-direction be the direction of the first piece.

I would suggest drawing a picture.

Let

m₁ = 2.5 kg be the mass of the first piece,

v₁ = 22 m/s be the speed of the first piece,

α = 0° be the angle the trajectory of the first piece forms with the positive x-axis,

m₂ = 3.2 kg be the mass of the second piece,

v₂ = 15.2 m/s be the speed of the second piece,

β = 55.4° be the angle the trajectory of the second piece forms with the positive x-axis,

m₃ be the mass of the third piece (to be computed),

v₃ = 33.5 m/s be the speed of the third piece,

γ be the angle the trajectory of the third piece forms with the positive x-axis (to be computed).

This is an exercise in conservation of momentum, which was 0 before the explosion.

We can express the conservation of momentum in two equations --

one the the x-axis and one for the y-axis.

Recall that a vector's projection on the x-axis is proportional to the cosine of its angle,

and a vector's projection on the y-axis is proportional to the sine of its angle.

m₁v₁cos(α) + m₂v₂cos(β) + m₃v₃cos(γ) = 0

m₁v₁sin(α) + m₂v₂sin(β) + m₃v₃sin(γ) = 0

The first equation represents the conservation of momentum along the x-axis, and

the second equation represents the conservation of momentum along the y-axis.

Rearrange the two equations to place the unknowns on the left

m₃v₃cos(γ) = -m₁v₁cos(α) - m₂v₂cos(β) (1)

m₃v₃sin(γ) = -m₁v₁sin(α) - m₂v₂sin(β) (2)

Question (a):

Divide equation (2) by equation (1)

tan(γ) = (m₁v₁sin(α) + m₂v₂sin(β)) / (m₁v₁cos(α) + m₂v₂cos(β)) (3)

Substituting actual numbers

tan(γ) = (2.5×22×sin(0°) + 3.2×15.2×sin(55.4°)) / (2.5×22×cos(0°) + 3.2×15.2×cos(55.4°)) = 0.4846

γ = 25.85° + k×180° (for any integer k) (4)

While (4) describes all the solutions to equation (3),

we are interested only in those solutions where 0° ≤ γ < 360°.

That means only for k = 0 or k = 1.

In addition, not all of those solutions are solutions to (1) and (2).

Equations (1) and (2) contain the additional information that

cos(γ) < 0 and sin(γ) < 0

That happens for γ satisfying 180° < γ < 270°.

Therefore k = 1, and

γ = 205.85°

Question (b)

Calculate m₃ from equation (2).

m₃ = (-m₁v₁sin(α) - m₂v₂sin(β))/(v₃sin(γ))

Substituting actual numbers, taking advantage of sin(0) = 0

m₃ = (-3.2×15.2×sin(55.4°)))/(33.5×sin(205.85°)) = 2.74 kg

Question (c)

In general, the efficiency of any process is the ratio between the amount of

"useful" output quantity divided by input quantity.

The quantity of interest may be energy, work, force.

What is designated as "useful" depends on the application.

In your example, there is only one quantity that can be discussed -- energy.

We know the kinetic energy of the pieces:

m₁v₁²/2 + m₂v₂²/2 + m₃v₃²/2

Besides this kinetic energy, the explosion yields other forms of energy, for example, heat.

If the objective of the explosion were the kinetic energy of the pieces, then

that would be designated as "useful" and the heat would be waste.

On the other hand, if the objective were the heat, then the kinetic energy

would be considered waste.

In order to calculate the efficiency, we would also need to know the energy

(chemical, nuclear, heat, pressure) of the original bomb.