Question

A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75

Answer 1

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Step-by-step explanation:

• Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.
• Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.
• Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:
• We can go with the x-axis first:

`p_(ox) = p_(fx) (1)`

⇒
`m_(tr) * v_(tr)= (m_(olds) + m_(tr)) * v_(fx) (2)`

• Replacing by the givens, we can find vfx as follows:

`v_(fx) = (m_(tr)*v_(tr) )/((m_(tr) + m_(olds)) ) = (4146kg*9.7m/s)/(2028kg+4146 kg) = 6.5 m/s (3)`

• We can repeat the process for the y-axis:

`p_(oy) = p_(fy) (4)`

⇒
`m_(olds) * v_(olds)= (m_(olds) + m_(tr)) * v_(fy) (5)`

• Replacing by the givens, we can find vfy as follows:

`v_(fy) = (m_(olds)*v_(olds) )/((m_(tr) + m_(olds)) ) = (2028kg*(-14.5)m/s)/(2028kg+4146 kg) = -4.8 m/s (6)`

• The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:

`v_(f) = \sqrt{v_(fx) ^(2) +v_(fy) ^(2) }} = \sqrt{(6.5m/s)^(2) +(-4.8m/s)^(2)} = 8.1 m/s (7)`

• In order to get the compass heading, we can apply the definition of tangent, as follows:

`(v_(fy) )/(v_(fx) ) = tg \theta (8)`

⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

⇒ θ = tg⁻¹ (-0.738) = -36.4º

• Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.