A block weighing 10 g, lying on a smooth table, is hit by a bullet weighing 2 g, flying at a speed of 60 m / s.

| November 19, 2020 | education

A block weighing 10 g, lying on a smooth table, is hit by a bullet weighing 2 g, flying at a speed of 60 m / s. How many millimeters will the bullet penetrate into the bar if the resistance to the movement of the bullet in the bar is 250N?

mp = 2 g = 2 * 10 ^ -3 kg.
mbr = 10 g = 10 * 10 ^ -3 kg.
V1 = 60 m / s.
F = 250 N.
l -?
Let’s use the law of conservation of momentum: mp * V1 = (mp + mbr) * V2.
Let’s find the speed of the bar with the bullet after the impact: V2 = mp * V1 / (mp + mbr).
V2 = 2 * 10 ^ -3 kg * 60 m / s / (2 * 10 ^ -3 kg + 10 * 10 ^ -3 kg) = 10 m / s.
The work of the bullet resistance force in the bar A is equal to the change in the kinetic energy of the bullet bar system ΔEk: A = ΔEk.
A = F * l.
l = A / F.
l = ΔEk / F.
ΔEk = mp * V1 ^ 2/2 – (mp + mbr) * V2 ^ 2/2.
ΔEk = 2 * 10 ^ -3 kg * (60 m / s) ^ 2/2 – (2 * 10 ^ -3 kg + 10 * 10 ^ -3 kg) * (10 m / s) ^ 2/2 = 3.6 J – 0.6 J = 3 J.
l = 3 J / 250 N = 0.012 m.
Answer: the bullet went deep into the block at a distance of l = 0.012 m.



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