Question

solution
(a) find the impulse delivered to the car.
calculate the initial and final momenta
of the car.
the impulse is just the difference
between the final and initial momenta.
(b) find the average force exerted on the car.
apply the impulse - momentum
theorem.
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remarks if the car doesnt rebound off the wall, the average force exerted on the car is smaller than
the value just calculated. with a final momentum of zero, the car undergoes a smaller change in
momentum.
question when a person is involved in a car accident, why is the likelihood of injury greater in a
head - on collision as opposed to being hit from behind? answer using the concepts of relative velocity,
momentum, and average force. (select all that apply.)
the change in momentum is greater in the head - on collision.
the average force on the driver is greater in the head - on collision.
the collapse of the crumple zone in the front of the car occurs only in the head - on collision.
the momentum of the driver relative to the ground is greater in a head - on collision.
the velocity of the driver relative to the ground is greater in a head - on collision.

Explanation:

• Relative velocity and momentum: In a head - on collision, the relative velocity of the driver with respect to the other vehicle (or object) is larger. Since momentum \(p = mv\), a larger relative velocity (assuming mass \(m\) is relatively constant for the driver) means a larger change in momentum \(\Delta p\).
• Impulse - momentum theorem (\(I=\Delta p = F_{av}\Delta t\)): Given the short collision time \(\Delta t\) (assuming similar collision times for head - on and rear - end collisions), a larger \(\Delta p\) (from the head - on collision) will result in a larger average force \(F_{av}\) (because \(F_{av}=\frac{\Delta p}{\Delta t}\)).

Answer:

• The average force on the driver is greater in the head - on collision.