Tank-killer rounds (fired by another tank) consist of a shell whose projectile is a long, slender (18.2 mm diameter by 31.7 cm long) cylindrical rod of depleted uranium of density 18.7 gm/cm3. Assume that the shell’s speed is 1550 m/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 speed is 280 m/s (in your direction, of course). If your counter-missile weighs 2.83 kg, how fast must it be going to achieve this result? Give the speed, not the velocity.
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Given:
Projectile mass (m1) = 18.7 gm/cm3 (density) * π * (0.091 cm)2 (radius) * 31.7 cm (length) = 2416.05 g = 2.41605 kg
Projectile velocity before collision (v1) = 1550 m/s
Counter-missile mass (m2) = 2.83 kg
Counter-missile velocity after collision (v2) = unknown
Combined lump velocity after collision (v) = 280 m/s
Using the principle of conservation of momentum:
m1 * v1 + m2 * v2 = (m1 + m2) * v
Substituting the given values:
(2.41605 kg) * (1550 m/s) + (2.83 kg) * v2 = (2.41605 kg + 2.83 kg) * 280 m/s
Simplifying the equation:
3742.7795 + 2.83 v2 = 1930.694
Rearranging the equation to isolate v2:
2.83 v2 = 1930.694 - 3742.7795
2.83 v2 = -1812.0855
Dividing both sides by 2.83 to solve for v2:
v2 = -1812.0855 / 2.83
v2 ≈ -639.98 m/s
Since the velocity should be a positive value, we take the magnitude of v2:
v2 ≈ 639.98 m/s
Therefore, the counter-missile must be moving at approximately 639.98 m/s to achieve the desired result.