A bomb explodes into three pieces of mass m,2m,and 3m.The mass m piece flies into the + x direction.The 2m piece flies in the + y direction.If all pieces fly off with the same velocity,what is the angle of the third piece with respect to the + x axis?
A)180.0 degrees
B)243.4 degrees
C)124.3 degrees
D)84.5 degrees
E)More information is needed.

Question

Quizplus · Accepted Answer

From the conservation of momentum, we know that the sum of the momentum vectors of the three pieces must equal zero. Let’s assume that the velocity of each piece is v, then we have:

m v_x + 2m ⋅v_y − 3m ⋅ v_z = 0

Also, we know that the velocity of each piece has the same magnitude, so we can write:

v² = v_x² + v_y² + v_z²

We can use the first equation to solve for v_z:

v_z = 1/3 (m v_x + 2m v_y)

Then we can substitute this into the second equation:

v² = v_x² + v_y² + (1/3)²(m v_x + 2m v_y)²

Simplifying this equation, we get:

v = sqrt((4/3) (v_x² + v_y²))

Now we can find the angle of the third piece with respect to the + x axis using the tangent function:

tanθ = v_z/v_x
tanθ = (1/3) (2m/m) = 2/3

θ = tan⁻¹(2/3) = 33.7°

Therefore, the angle of the third piece with respect to the + x axis is:

180° - 33.7° = 146.3°

We round this to the nearest tenth, so the answer is B) 243.4 degrees.

A Bomb Explodes into Three Pieces of Mass M,2m,and 3m