Center of Mass, Impulse, Momentum, and Collisions

Question

You are driving a truck of mass 5 650 kg at 13 m/s. You accidentally hit a parked car. The car and the passengers have a total mass of 984 kg. (a) What is the speed of the car and the truck after the impact assuming that they are locked together? (b) If the impact lasts for 0.5 s, what force is experienced by an 80 kg passenger? Use GUESS METHOD

Answer 1

Let's think of this in terms of the GUESS method:

GIVEN:

Mass of the truck: m_truck = 5650 kg

Initial velocity of the truck (before the collison): v_truck,i = 13 m/s

Mass of the car with passengers: m_car+pass = 984 kg

Initial velocity of the car with passengers (before the collision): v_car+pass,i = 0 m/s (since the car was parked)

Truck and the car (with passengers) have the same final velocity (since the car and the truck are stuck together after the collision)

Mass of a passenger in the parked car: m_pass = 80 kg

Initial velocity of the passenger (before the collision): v_pass,i = 0 m/s (since the car was parked)

Time of impact: t_impact = 0.5 s

Passenger has the same final velocity as the car (and therefore the truck), since the passenger is inside the car.

UNKNOWN:

Final velocity of the car, passenger, and truck after collision: v_f (NOTE: In general, we would need separate unknowns for the car, truck, and passenger. However, since we know from our given that their final velocities are all the same, I'll just assign one unknown to make things simpler)

Average force experienced by the passenger: F_avg,pass

EQUATIONS:

In an isolated system, momentum is conserved. Therefore, we can set up a momentum equation for the collision: m_truckv_truck,i + m_car+passv_car+pass,i = (m_truck + m_car+pass)v_f

We also know that the change in momentum experienced by a body is equal to the Impulse. We have for the passenger:

Impulse = F_avg,pass t_impact

Change in Momentum = m_pass(v_f - v_pass,i)

Equating gives another equation that we can use: F_avg,pass t_impact = m_pass(v_f - v_pass,i)

For the last part, I'll consider parts a) and b) of the problem separately:

PART A SUBSTITUTE:

From the momentum equation: 5650 (13) + 984 (0) = (5650+984)v_f

PART A SOLVE:

5650 (13) + 984 (0) = (5650+984)v_f

5650 (13) = (5650+984)v_f

v_f = 5650 (13)/(5650+984) ≈ 11.07 m/s

v_f ≈ 11.07 m/s

PART B SUBSTITUTE (Using result from part A):

From the impulse equation:

F_avg,pass (0.5) = 80 (11.07 - 0)

PART B SOLVE:

F_avg,pass (0.5) ≈ 80 (11.07 - 0) (this is approximate since I'm using a rounded value for the final velocity)

F_avg,pass ≈ 80 (11.07)/(0.5)

F_avg,pass ≈ 1771.2 N