Most of us know intuitively that in a head-on collision between a large dump tru

Question

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck in only dented. This idea of unequal forces, of course, is false. Newton's third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 6.0 m/s and that they undergo a perfectly inelastic head-on collision. (In an inelastic collision, the two objects move together as one object after the collision.) Each driver has a mass of 80.0 kg. Including the drivers, the total vehicle masses are 880 kg for the car and 4080 kg for the truck. The collision time is 0.120 s. Choose coordinates such that the truck is initially moving in the positive x direction, and the car is initially moving in the negative x direction.

(a) What is the total x-component of momentum BEFORE the collision?

(b) What is the x-component of the CENTER-OF-MASS velocity BEFORE the collision?

(c) What is the total x-component of momentum AFTER the collision?

(d) What is the x-component of the final velocity of the combined truck-car wreck?

(e) What impulse did the truck receive from the car during the collision? (Sign matters!)

(f) What impulse did the car receive from the truck during the collision? (Sign matters!)

(g) What is the average force on the truck from the car during the collision? (Sign matters!)

(h) What is the average force on the car from the truck during the collision? (Sign matters!)

(i) What impulse did the truck driver experience from his seatbelt? (Sign matters!)

(j) What impulse did the car driver experience from his seatbelt? (Sign matters!)

(k) What is the average force on the truck driver from the seatbelt? (Sign matters!)

(l) What is the average force on the car driver from the seatbelt? (Sign matters!)

Explanation / Answer

1. What do you think the difference is between the (e) and (i) questions?

(e) What impulse did the truck receive from the car during the collision? (Sign matters!)
(i) What impulse did the truck driver experience from his seatbelt? (Sign matters!)

The truck driver is initially moving with the truck, so he has the same initial velocity Vi; and he also has the same final velocity Vf, because the belt is attached to the truck. So the Vf - Vi = ?V is the same. The impulse J is given by the same formula:

(1) J = ?p = ?(m * v) = m * ?v [when the mass doesn't change];

so the only difference between the truck and the driver is their masses.

So for (i) use the same formula as (e) except use the mass of the driver.

Note: The maximum number of significant digits in the given values is 3, so your calculations should reflect that. Specifically, for (e):

mass of truck (final velocity-initial velocity) = 4060 (5.85-9) = -1.28 kg*m/s


2. Similarly, for (j), (k), and (l), notice the similarities with (f), (g), and (h).

Note, however: In (g) you write:

Impulse = Force (time) -12789 = F(0.08) F= -159862.5 N

The "impulse / momentum" theorem says that (see Source 1):

(2) J = ?p = F * ?t

Therefore,

(3) F = ?p / ?t

= -1.28 / 0.080

= -1.60e5 N

I know you got the same answer, but it looks like you're multiplying F by 0.08 instead of dividing the change in momentum by 0.080.

Mass of the car with driver, m1 = 810 kg

Mass of the truck with driver, m2 = 4010 kg

Initial velocity of the car, v1 = 10 m/s

Initial velocity of the truck, v2 = - 10 m/s

In the perfectly inelastic collision,

Common velocity of the system, v = ( m1 v1 + m2 v2 ) / (m1+m2)

                                                    = (8100 - 40100)/(810+4010)

                                                    = - 6.64 m/s

----------------------------------------------------------------------

Mass of each driver, M = 90 kg

Collision time, t = 0.140 s

(a)

Force exerted by belt on the truck driver:

Ftruck= M (v-v2)/t

        = 90 * (-6.64+10)/0.140

        =2.16 x 10^3 N

(b)

Force exerted by belt on the car driver:

Fcar = M (v-v1)/t

        = 90 * (-6.64-10)/0.140

        =1.07 x 10^4 N

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