Answer to Question #141200 in Mechanics | Relativity for Daarren

Question

A 5.00-g bullet moving with an initial speed of v1 = 400 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with a force constant 900 N/m. The block moves 5.00 cm to the right after impact before being brought to rest by the spring. Find (a) the speed at which the bullet emerges from the block assuming that the energy is conserved.

Answer 1

The total momentum of the isolated system of the two particles (bullet and block) must remain constant:

$$m_1 v_{1b}+m_2 v_{2b}=m_1 v_{1a}+m_2 v_{2a} \qquad(1)$$

m₁ = 5g = 0.005 kg - bullet mass

m₂ = 1 kg - block mass

v_1b = 400 m/s - bullet velocity before impact

v_2b = 0 m/s - block velocity before impact

v_1a - bullet velocity after impact

v_2a - block velocity after impact

$$v_{1a}=\frac {m_1 v_{1b}-m_2 v_{2a}}{m_1}=v_{1b}-\frac{m_2}{m_1}v_{2a}\qquad(2)$$

The total energy of the isolated system (spring and block) must remain constant:

$$K_b+U_b=K_a+U_a\qquad(3)$$

The potential energy of the spring (U_b) before compression and the kinetic energy of the block (K_a) after compression are equal to zero.

Kinetic energy of the block before compression (after impact):

$$K_b=\frac{m_2 v_{2b}^2}{2} \qquad(4)$$

Potential spring energy after compression (from Hooke's law):

$$U_a=\frac{kx^2}{2}\qquad(5)$$

From (3),(4) and (5):

$$\frac{m_2 v_{2b}^2}{2}=\frac{kx^2}{2}$$ $$v_{2b}=x\sqrt{\frac{k}{m_2}}\qquad(6)$$

k = 900 N/m - constant factor characteristic of the spring (i.e.,its stiffness)

x = 5 cm = 0.05 m - deformation of the spring

So, from (2) and (6):

$$v_{1a}=v_{1b}-\frac{m_2}{m_1}x\sqrt{\frac{k}{m_2}}=v_{1b}-\frac{x\sqrt{m_2k}}{m_1}$$ $$v_{1a}=400\frac{m}{s}-\frac{\sqrt{1kg\cdot900\frac{N}{m}}}{0.005kg}0.05m=100\frac{m}{s}$$

Answer: the speed at which the bullet leaves the block is 100 m/s