# A plasticine bullet with a mass of 9 g flying horizontally at a speed of 20 m / s hits a load motionlessly hanging

**A plasticine bullet with a mass of 9 g flying horizontally at a speed of 20 m / s hits a load motionlessly hanging on a thread 40 cm long, as a result of which the load with a bullet adhered to it vibrates. The maximum angle of deflection of the thread from the vertical is 60 (degrees). What is the weight of the cargo?**

given

v1 = 20 m / s

m1 = 9 grams = 0.009 kg

L = 40cm = 0.4m

angle a = 60 degrees

find the mass of the load M2

solution

consider a right-angled triangle, where L is the hypotenuse, then through the cosine we find the height to which the load with the bullet rises:

H = cos 60 * L = 1/2 * 0.4 m = 0.2 m

According to the law of conservation of energy

(m1 + m2) * V2 ^ 2/2 = (m1 + m2) * g * h

from here

V2 ^ 2/2 = g * h

V2 ^ 2 = g * h * 2

that is, V2 = √ (2 * g * h)

V2 = √ (2 * 10 * 0.2) m / s = √ (4) = 2 m / s

the law of conservation of momentum of the system is equal to:

m1 * V1 + 0 = (m1 + m2) * V2

m1 * V1 = V2 * m1 + m2 * V2

m1 * V1 – V2 * m1 = m2 * V2

hence m2 = (m1 * V1) / V2 – m1

m2 = 0.009 * 20/2 – 0.009 = 0.081 kg = 81 grams

answer: 0.081 kg = 81 grams