Question

На горизонтальной пружине укреплено тело массой 10кг, лежащее на абсолютно гладком столе. В тело попадает и застревает в нем пуля массой 10г, летящая 500 м/с направленной вдоль оси пружины. Амплитуда возникших при этом колебаний равна 0.1м. определите период возникших колебаний

Answer 1

M*V1=(m1+m2)*Vm
Vm=m*V1/(m1+m2)=0,01*500/10=0,5 м/с
Vm=xm*w
w=Vm/xm=5/0,1=5 рад/с
T=2*pi/w=6,28/5=1,26 c

Answer 2

Calculation of the Period of Oscillations

To determine the period of the oscillations that occur after the bullet gets embedded in the body, we can use the principle of conservation of momentum. The initial momentum of the system is equal to the final momentum of the system.

The initial momentum of the system is given by the product of the mass of the bullet (m) and its velocity (v), while the final momentum is given by the product of the combined mass of the body and the bullet (M + m) and the velocity of the center of mass of the system (V).

According to the principle of conservation of momentum, we have:

m * v = (M + m) * V

Given that the mass of the body (M) is 10 kg, the mass of the bullet (m) is 10 g (or 0.01 kg), and the velocity of the bullet (v) is 500 m/s, we can substitute these values into the equation:

0.01 kg * 500 m/s = (10 kg + 0.01 kg) * V

Simplifying the equation:

5 kg m/s = 10.01 kg * V

V = 5 kg m/s / 10.01 kg

V ≈ 0.4995 m/s

Now, we can calculate the period of the oscillations using the formula:

T = 2π * √(m / (M + m)) / V

Substituting the values:

T = 2π * √(0.01 kg / (10 kg + 0.01 kg)) / 0.4995 m/s

T ≈ 2π * √(0.01 kg / 10.01 kg) / 0.4995 m/s

T ≈ 2π * √(0.000999) / 0.4995 m/s

T ≈ 2π * 0.03162 / 0.4995 m/s

T ≈ 0.1989 s

Therefore, the period of the oscillations is approximately 0.1989 seconds.