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

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

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:

       $v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)$

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

       $v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)$

• 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:

       $tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)$

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E