**Worked examples**

In accordance with the principle of conservation of energy, the whole of this potential energy becomes transferred to kinetic energy when the stone reaches the ground again. Hence kinetic energy on reaching the ground = 56.25 J. Answer. P.E. at greatest height During a shunting operation, a truck of total mass 15 metric tonnes (t) moving at 1 mis, collides with a stationary truck of mass 10 t. If the two trucks are automatically connected so that they move off together, find their velocity. Also calculate the kinetic energy of the trucks: (a) before; (b) after collision. Explain why these are not equal. (1 t = 1000 kg.) By the principle of conservation of omentum, momentum before collision = momentum after collision Let v = common velocity after collision, then using t mls units of momentum, (15 x 1) + (10 x 0) = (15 + 10) x v 15

or v = 25 = 0.6 mls

K.E. = tmv2 (m in kg; v in m/s)

= t x 15000 x 12 = 7500 J

= t x 25000 x 0.62 = 4500 J

Answer. Velocity before collision = 0.6 mls

K.E. before collision = 7.5 kJ

K.E. after collision = 4.5 kJ

Using the formula

K.E. before collision

K.E. after collision

In accordance with the principle of conservation of energy, the total energy after collision is the same as that before. Before collision the whole of the energy is kinetic in the moving truck, but when collision occurs part of this becomes transferred to internal energy in both trucks (k.e. and p.e. of molecules, see page 76) and part into sound energy (k.e. and p.e. of air

molecules). The remainder is left as mechanical kinetic energy in both trucks. Consequently, mechanical kinetic energy after collision is less than mechanical kinetic energy before collision.

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