Question

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of m A = 17.0 kg and an initial velocity of v 0A = 8.00 m/s, due east. Object B, however has a mass of m B = 29.0 kg and an initial velocity of v 0B = 5.00 m/s, due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.

Answer 1

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Step-by-step explanation:

• Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
• If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

`p_(ox) = p_(oAx) + p_(oBx) (1)`

• We can do exactly the same for the initial momentum along the y-axis:

`p_(oy) = p_(oAy) + p_(oBy) (2)`

• The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

`p_(fx) = (m_(A) + m_(B) ) * v_(fx) (3)`

• We can repeat the process for the y-axis, as follows:

`p_(fy) = (m_(A) + m_(B) ) * v_(fy) (4)`

• Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

`v_(fx) = (p_(oAx))/((m_(A)+ m_(B))) = (m_(A)*v_(oAx) )/((m_(A)+ m_(B))) =(17.0kg*8.00m/s)/(46.0kg) = 2.96 m/s (5)`

• In the same way, we can find the component of the final momentum along the y-axis, as follows:

`v_(fy) = (p_(oBy))/((m_(A)+ m_(B))) = (m_(B)*v_(oBy) )/((m_(A)+ m_(B))) =(29.0kg*5.00m/s)/(46.0kg) = 3.15 m/s (6)`

• With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

`v_(f) = \sqrt{v_(fx) ^(2) + v_(fy) ^(2)} = \sqrt{(2.96m/s)^(2) + (3.15m/s)^(2)} = 4.32 m/s (7)`

• The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

`p_(f) = (m_(A) + m_(B))* v_(f) = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)`

• Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
• We can find this angle applying the definition of tangent of an angle, as follows:

`tg \theta = (v_(fy))/(v_(fx)) = (3.15 m/s)/(2.96m/s) = 1.06 (9)`

⇒ θ = tg⁻¹ (1.06) = 46.8º N of E