A 1600 kg white sedan going 70 mph is involved in a T-bone collision with a 1000 kg red coupe traveling at?

Question

A 1600 kg white sedan going 70 mph is involved in a T-bone collision with a 1000 kg red coupe traveling at 35 mph. The collision is completely inelastic and lasts for 0.10 seconds. Assume that the sedan is going in the negative x direction and the coupe is traveling in the positive y direction immediately before they collide. All masses above include any passengers and cargo.

a) What is the velocity (magnitude and direction) of the wreckage immediately after the collision, in mph?

b) What is the magnitude of the average force experienced by a 80 kg dummy in the sedan during the collision, in pounds? (You may assume that the final velocity of the dummy is the same as the rest of the wreckage.)

c) How much kinetic energy was lost by the cars (including all passengers and cargo) in the collision, in Joules?

Woodsman · Accepted Answer

initial momentum p = -1600kg * 70mph i + 1000kg * 35mph j
 = final momentum p = -112000kg·mh i + 35000kg·mph j
Divide by total mass (2600kg) to get velocity:
v = -43mph i + 13.5mph  j
mag v = √(43² + 13.5²) mph = 45 mph ← magnitude
Θ = arctan(13.5 / -43) = -17.3º + 180º = 163º ccw from +x axis

(b) initially, sedan v = -70mph i
Later, sedan v = -43mph i + 13.5mph j
Δv = 27mph i + 13.5mph j
mag Δv = √(27² + 13.5²) mph = 30.2mph * 22ft/s / 15mph = 44.3 ft/s
so the acceleration a = Δv / Δt = 44.3ft/s / 0.1s = 443 ft/s²
mass m = 80kg * 2.2lbm/kg * 1slug/32lbm = 5.5 slugs
so the force F = ma = 5.5slugs * 443ft/s² = 2435 lbf

(c) convert velocities to m/s
70 mph ≈ 31.3 m/s
35 mph ≈ 15.65 m/s
45 mph ≈ 20.1 m/s
initial KE = ½ * 1600kg * (31.3m/s)² + ½ * 1000kg * (15.65m/s)²
final KE = ½ * 2600kg * (20.1m/s)²
loss = 381 kJ

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A 1600 kg white sedan going 70 mph is involved in a T-bone collision with a 1000 kg red coupe traveling at?

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