Find out also the loss of energy through impact:

Example

1. A bullet weighing 0.1 newton horizontally fired with a velocity of 80 metres/sec hits a wooden block, weighing 0.9 Newton. This penetrates into the block through a distance of 8 cms while the latter is held firmly in a clamp. Find out the average resistance offered by the wooden block to the penetration of the bullet.

2. If the block were free to move on a smooth horizontal plane, find out the penetration of the bullet. Find out also the loss of energy through impact.

Solution

Consider the average resistance of wooden block to the penetration of the bullet be R Newtons.

 (i)        While block is held fixed in a clamp, the whole of Kinetic Energy (KE) of the bullet is lost in doing work of penetration of 8 cm (that means 0.08 metres).

∴ Work done by R = KE of bullet

R × 0.08 = (0.1 × 80 × 80) / (2 × 9.8)

∴ R = 408.16 N.

 (ii)       While the block is free to move horizontally, Consider V_c be the common velocity after impact

∴ (0.1 + 0.9) × Vc / g = (0.1 × 80) /g

∴ Vc  = 8 metres / sec .

Entire KE of the combined mass after impact

=  1 × 8² /2 × 9.8      = 3.27 N. m.                             ---------- (1)

Total KE of the system before impact

= (0.1 × 80 × 80 )/2 g   = 32.65 N-m-------------- (2)

∴ Energy lost in impact (i.e. in causing penetration)

                        = (1) - (2) = 29.38 N.m         ------------ (3)

Neglecting other losses like heat, sound etc. during the process of penetration, if we assume that resistance R absorbs this change in KE given in Eq. (3).

             R × l = 29.38

∴ l = 29.38 /408.16     = 0.072 m.

 Note down that penetration of 7.2 cm under case (ii) is less than that case (i).

N. B.    If the wooden block is free to move on a horizontal surface, it shall continue to move along a constant velocity V_c. Though, if R′ is a resistance offered so that it comes to rest through a distance s′, then

R′ × s′ = KE of the combined mass after the impact.

= 1 ×( V_c)² / 2 g