3.7 Collisions and Impulse

In solving problems regarding the collision of objects, the Law of Conservation of Momentum is often restated as:

The subscripts 1 and 2 refer to two different objects, and the prime symbol (¢ ) signifies a quantity after the collision. Thus m1v1 represents the momentum of object 1 before the collision, while m1¢ v1¢ represents object 1’s momentum after the collision.

Impulse, the change in momentum of an object, is defined as a force applied through time. A mathematical translation follows:

The unit of impulse is N× s. This unit is equal to the SI unit for momentum, k× m/s, as shown below:

The relationship between energy, momentum, and impulse is illustrated in the following situation.

Calculate (a) the impulse experienced when a 70-kg person lands on firm ground after jumping from a height of 5.0 m (ignore air friction). Then estimate the average force exerted on the person’s feet by the ground if the landing is (b) stiff-legged, and (c) with bent legs. In the former case, assume the body moves 1.0 cm during impact, and in the second case, assume the legs bend 50 cm.

Solution (a): The force exerted by the person on the ground (and by the ground on the person) is unknown, and thus impulse cannot be directly calculated. However, we can use the fact that impulse is equal to the change in momentum of the object. First we will calculate the person’s velocity at the moment he lands on the ground:

As the person strikes the ground, the momentum is quickly brought to zero. The impulse on the person is thus:

Solution (b): In coming to rest, the body accelerates from 9.9 m/s to zero in a distance, d = 1.0 cm = 1.0 x 10-2 m. The average velocity during this period is (9.9 m/s + 0 m/s)/2 = 5.0 m/s, so the time of the collision is

Since the impulse is FD t = 690 N× s, and D t = 2.0 x 10-3 s, the average force, Favg, is

By Newton’s Second Law, Favg is the average upward force exerted on the man’s legs by the ground, Fgnd, minus the downward force of gravity, mg:

Solution (c): This case is similar to that in part (b), except d = 0.50 m. Thus, D t = (0.5 m)/(5.0 m/s) = 0.1 s, and