Impulse and Impact

Back to main index

Worked Examples

Impact of inelastic bodies

When two inelastic bodies collide they remain together. They show no inclination to return to their original shape after the collision.

An example of this may be two railway carriages that collide and become coupled on impact.

Problems of this type may be solved by the principle of conservation of linear momentum.

(this must be applied in the same direction)

Although momentum is conserved, it is important to realise that energy is always lost in an inelastic collision (it is converted from mechanical energy to some other form such as heat, light or sound.)

Back to the top

Worked Example 4.3

A railway wagon of mass 20 tonnes travelling at 1.5m/s collides with another of mass 30 tonnes travelling in the opposite direction at 0.5m/s. The wagons become coupled on impact. Find:

  • their common velocity after impact
  • the loss of kinetic energy.

  • Solution

    a)

    Note the negative sign for the second wagon as positive is taken as the direction of velocity of the 20 tonne wagon.

    After the impact, is the common velocity is V then the momentum will be (20000 + 30000)V

    using the conservation of linear momentum

    This is positive, so it is in the original direction of the 20 tonne wagon.

    b)

    Back to the top

    Worked Example 4.4

    A pile-driver of mass 2.5 tonnes drives a pile of mass 500 kg vertically into the ground. The driver falls freely a vertical distance of 2m before hitting the pile and there is no rebound. Each blow of the drive moves the pile down 0.2m.

    What is the average value of resistance of the ground to penetration?

    Solution

    The velocity of the pile-driver just before it hits the pile can be found using Equation 1.4

    u = 0.0, a = 9.81m/s2, s = 2m

    The momentum just before impact is thus

    Since there is no rebound, the pile and driver have the same velocity after impact. So we can write this expression for momentum after impact if the common velocity is V:

    So by the principle of conservation of momentum

    The pile and driver are now brought to rest by the deceleration force of the ground in 0.2m. we can find this deceleration using Equation 1.4

    u = 5.22 m/s, s = 0.2m, v = 0.0

    Now the retarding force is given by

    the ground resistance, R, is the sum of this retarding force and the weight of the pile and driver

    Back to the top