1.) Tank-killer rounds (fired by another tank) consist of a shell whose projecti

Question

1.) Tank-killer rounds (fired by another tank) consist of a shell whose projectile is a long, slender (19.1 mm diameter by 31.2 cm long) rod of depleted uranium of density 18.7 g/cm3. Assume that the shell’s speed is 1.55 km/s. You have developed a defense: you fire a counter-missile that hits the incoming projectile head-on and sticks to it. The object is to reduce the velocity of the combined lump to less than the minimum value that can penetrate your armor. In your case, that velocity is 280 m/s (in your direction, of course). If your counter-missile weighs 3.06 kg, how fast must it be going to achieve this result?
not = 384.13

Explanation / Answer

       Le the mass of tank killer shell be m1 and its initial speed be v1 while the defence projectile have a mass of m2 and initial speed be v2, and the combined fina speed is V, then according to the principle of conservation of linear momentum    m1 * v1 + m2 * v2    =   (m1 +   m2) * V    m1   =   volume of cylindrical shell * density          =    * r2 * L * d          =   3.14 * (19.1 * 10-3 / 2)2 * 31.2 * 10-2 * 18.7 * 103                                  (as 1 mm = 10-3 m, 1 cm = 10-2 m, 1 g / cm3   =   103   kg/m3)          =   1.671   kg    given      v1   =   1.55 km/s   =   1550   m/s                   m2 =   3.06 kg                   V   =   280   m/s    substituting values       1.671 * 1550   + 3.06 * v2    =    (1.671 + 3.06) * 280       2590.05   +   3.06 * v2   =   1324.68    velocity of defensive projectile      v2    =   (1324.68 - 2590.05) / 3.06                                                                   =   - 413.52   m/s    - ve sign indicates that the defensive projectile must be hit the incoming shell head on, and hence can be dropped.

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