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 10.0 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 100.0 kg. Including the drivers, the total vehicle masses are 900 kg for the car and 4100 kg for the truck. The collision time is 0.090 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!)
To answer these questions, we need to apply the principles of conservation of momentum and the concept of impulse.
(a) The total x-component of momentum before the collision is the sum of the momenta of the truck and the car. The momentum is given by the product of mass (m) and velocity (v). Therefore, the momentum of the truck is (4100 kg)(10.0 m/s) = 41000 kg·m/s in the positive x-direction, and the momentum of the car is (900 kg)(-10.0 m/s) = -9000 kg·m/s in the negative x-direction. The total momentum is the sum of these values, so (41000 - 9000) kg·m/s = 32000 kg·m/s in the positive x-direction.
(b) The center-of-mass velocity is the total momentum divided by the total mass. The total mass is the sum of the masses of the truck and the car. Therefore, the center-of-mass velocity is (32000 kg·m/s) / (4100 kg + 900 kg) = 7.0968 m/s in the positive x-direction.
(c) The total x-component of momentum after the collision is the same as before the collision because momentum is conserved. Therefore, the total x-component of momentum after the collision is 32000 kg·m/s in the positive x-direction.
(d) After the collision, the truck and the car move together as one object. The final velocity of the combined truck-car wreck can be found by dividing the total momentum after the collision by the total mass of the combined system. The total mass of the system is 4100 kg + 900 kg = 5000 kg. Therefore, the final velocity of the combined truck-car wreck is (32000 kg·m/s) / (5000 kg) = 6.4 m/s in the positive x-direction.
(e) The impulse experienced by an object is equal to the change in momentum it experiences during a collision. Since the truck and the car move together after the collision, the change in momentum of the truck is 32000 kg·m/s - 41000 kg·m/s = -9000 kg·m/s in the positive x-direction. Therefore, the impulse received by the truck from the car during the collision is -9000 kg·m/s.
(f) Similarly, the change in momentum of the car is 32000 kg·m/s - (-9000 kg·m/s) = 41000 kg·m/s in the negative x-direction. Therefore, the impulse received by the car from the truck during the collision is 41000 kg·m/s.
(g) The average force on an object can be calculated by dividing the impulse it receives by the time interval over which the impulse is applied. In this case, the collision time is given as 0.090 s. Therefore, the average force on the truck from the car during the collision is (-9000 kg·m/s) / (0.090 s) = -100000 N in the positive x-direction.
(h) Similarly, the average force on the car from the truck during the collision is (41000 kg·m/s) / (0.090 s) = 455556 N in the negative x-direction.
(i) The impulse experienced by the truck driver from his seatbelt is equal to the change in momentum of the truck divided by the collision time. The change in momentum of the truck is -9000 kg·m/s, and the collision time is 0.090 s. Therefore, the impulse experienced by the truck driver from his seatbelt is (-9000 kg·m/s) / (0.090 s) = -100000 N·s.
(j) Similarly, the impulse experienced by the car driver from his seatbelt is 41000 kg·m/s / 0.090 s = 455556 N·s.
(k) The average force on the truck driver from the seatbelt can be calculated by dividing the impulse experienced by the truck driver by the collision time. Therefore, the average force on the truck driver from the seatbelt is (-9000 kg·m/s) / (0.090 s) = -100000 N.
(l) Similarly, the average force on the car driver from the seatbelt is (41000 kg·m/s) / (0.090 s) = 455556 N.
These calculations help us understand the quantities involved in the collision, the forces experienced by the objects, and the impulses experienced by the drivers and their seatbelts.